Quadratic Equation Notes

QUADRATIC EQUATION CONTENTS  Basic Concept  Value of a Quadratic Polynomial  Zeros of a Quadratic Polynomial  Quadratic Equation & Its Roots   Solving a Quadratic Equation By Factorisation Solving a quadratic Equation By Completing the square   Solving a Quadratic Equation By Using Quadratic Formula Nature or Character of the Roots of a Quadratic Equation  Sum and Product of The Roots  To Construct A Quadratic Equation Whose Roots Are Given   Equation Reducible to Quadratic Equations Problems on Quadratic Equations Every algebraic polynomial of second degree is called a quadratic polynomial. For Example : (i) 3x2 + 5x + 7 (ii) 8x2 – 6x (iii) 5x2 – 7 (iv) 2 x2 + 6x – 3 The general form of quadratic polynomial is ax2 + bx + c; where a, b, c are real numbers, a  0 and x is variable. For a particular quadratic polynomial the values of a, b, c are constant and for this reason a, b and c are also called real constants. For example, in quadratic polynomial 3x2 – 5x + 8; 3, – 5 and 8 are constant where as x is variable.  VALUE OF A QUADRATIC POLYNOMIAL The value of a quadratic polynomial ax2 + bx + c (i) at x =  is a()2 + b() + c = a2 + b + c (ii) at x =  is a2 + b + c (iii) at x = 5 is a (5)2 + b(5) + c = 25a2 + 5a + c In the same way : (i) Value of 5x2 – 3x + 4 at x = 2 is = 5(2)2 – 3(2) + 4 = 20 – 6 + 4 = 18 (ii) Value of x2 – 8x – 15 at x = – 1 is = (–1)2 – 8 (–1) – 15 = 1 + 8 – 15 = – 6 2 (iii) Value of 7x2 – 4 at x =  BASIC CONCEPT 2 2 is = 7   – 4 3 3 4 –4 9 Ex.2 Find the value of m, if x = 2 is a zero of quadratic polynomial 3x2 – mx + 4. 28  36 8 = 9 9 Sol. Since, x = 2 is a zero of 3x2 – mx + 4 =7× =  3(2)2 – m × 2 + 4 = 0  12 – 2m + 4 = 0, i.e., m = 8.  ZEROS OF A QUADRATIC POLYNOMIAL The value of the polynomial x2 – 7x + 10 at :  QUADRATIC EQUATION & IT’s ROOTS (i) x = 1 is (1)2 – 7 × 1 + 10 = 1 – 7 + 10 = 4 (ii) x = 2 is (2)2 Since, ax2 + bx + c, a  0 is a quadratic polynomial, ax2 + bx + c = 0, a  0 is called a quadratic equation. – 7 × 2 + 10 = 4 – 14 + 10 = 0 (iii) x = 3 is (3)2 – 7 × 3 + 10 = 9 – 21 + 10 = – 2  (iv) x = 5 is (5)2 – 7 × 5 + 10 = 25 – 35 + 10 = 0 It is observed here that for x = 2 and x = 5; the value of polynomial x2 –7x + 10 is zero. These two values of x are called zeros of the polynomial. Thus, if for x = , where  is a real number, the value of given quadratic polynomial is zero; the real number  is called zero of the quadratic polynomial. EXAMPLES  Ex.1 (i) x = 3 is a zero of quadratic polynomial x2 – 2x – 3. (ii) x = – 2 is a zero of quadratic polynomial 3x2 + 7x + 2. (iii) x = 4 is not a zero of quadratic polynomial 2x2 – 7x – 5. (3)2 (ii) 5x2 – 7x = 0 is a quadratic equation. (iii) 5x2 + 2 = 0 is a quadratic equation, but (iv) –7x + 2 = 0 is not a quadratic equation.  EXAMPLES  Ex.3 Which of the following are quadratic equations, give reason : (i) x2 – 8x + 6 = 0 Show that : Sol. (i) The value of (i) –x2 – 7x + 2= 0 is a quadratic equation, as –x2 – 7x + 2 is a quadratic polynomial. x2 – 2x – 3 at x = 3 is (iii) 2x + 5 = x2 x (iv) x2 + 2 =3 x2 Sol. (i) Since, x2 – 8x + 6 is a quadratic polynomial  x2 – 8x + 6 = 0 is a quadratic equation. –2×3–3=9–6–3=0  x = 3 is a zero of quadratic polynomial – 2x – 3. (ii) 3x2 – 4 = 0 x2 (ii) The value of 3x2 + 7x + 2 at x = – 2 is 3(–2)2 + 7 (–2) + 2 = 12 – 14 + 2 = 0  x = – 2 is a zero of quadratic polynomial 3x2 + 7x + 2 (iii) The value of 2x2 – 7x – 5 at x = 4 is 2(4)2 – 7(4) – 5 = 32 – 28 – 5 = – 1  0  x = 4 is not a zero of quadratic polynomial 2x2 – 7x – 5. (ii) 3x2 – 4 = 0 is a quadratic equation. (iii) 2x + 5 = x3 x  2x2 + 5 = x3  x3 – 2x2 – 5 = 0; which is cubic and not a quadratic equation. (iv) x2 + 2 =3 x2  x4 + 2 = 2x2  x4 – 2x2 + 2 = 0; which is biquadratic and not a quadratic equation. Ex.4 In the same way : In each of the following, determine whether the given values are solutions (roots) of the equation or not : 3x2 – 5x + 2 = 0 3x2 – 3x – 2x + 2 = 0 [Factorising L.H.S.] (i) 3x2 – 2x – 1 = 0; x = 1 (x – 1) (3x – 2) = 0 (ii) x2 + 6x + 5 = 0; x = – 1, x = – 5 (iii) x2 + 2 x – 4 = 0; x = 2,x=–2 2 Sol. (i) Value of 3x2 – 2x – 1 at x = 1 is i.e., x–1=0 or 3x – 2 = 0  x=1 or x= 2 ; 3 which is the solution of given quadratic equation. 3(1)2 – 2(1) – 1 = 3 – 2 – 1 = 0 = RHS  x = 1 is a solution of the given equation. In order to solve the given Quadratic Equation: 1. Clear the fractions and brackets, if given. (ii) For x = – 1, L.H.S. = (–1)2 + 6 (–1) + 5 2. By transfering each term to the left hand side; express the given equation as ; ax2 + bx + c = 0 or a + bx + cx2 = 0 = 1 – 6 + 5 = 0 = R.H.S.  x = – 1 is a solution of the given equation For x = – 5, L.H.S. = (–5)2 + 6(–5) + 5 3. Factorise left hand side of the equation obtained (the right hand side being zero). = 25 – 30 + 5 = 0 = R.H.S.  x = – 5 is a solution of the given equation. (iii) For x = 2 , L.H.S. = x2 + = ( 2 )2 +  2x–4 2( 2)–4  EXAMPLES  Ex.5 Solve : =2+2–4=0 (i) x2 + 3x – 18 = 0 = R.H.S. (iii) 2x2 + ax – a2 = x = 2 is a solution of the given equation For x = – 2 2 , L.H.S. = (–2 2 )2 + Sol. (ii) (x – 4) (5x + 2) = 0 0; where ‘a’ is a real number. (i) x2 + 3x – 18 = 0 x2 + 6x – 3x – 18 = 0 x(x + 6) – 3(x + 6) = 0 2 ×–2 2 –4 = 4 × 2 – 2 × 2 – 4 = 0 R.H.S.  4. By putting each factor equal to zero; solve it. x = – 2 2 is a solution of the given equation. SOLVING A QUADRATIC EQUATION BY FACTORISATION Since, 3x2 – 5x + 2 is a quadratic polynomial; 3x2 – 5x + 2 = 0 is a quadratic equation. Also, 3x2 – 5x + 2 = 3x2 – 3x – 2x + 2 [Factorising] = 3x (x – 1) – 2(x – 1) = (x – 1) (3x – 2) i.e., (x + 6) (x – 3) = 0  x + 6 = 0 or x – 3 = 0  x = – 6 or x = 3 Roots of the given equation are : – 6 and 3 (ii) (x – 4) (5x + 2) = 0x – 4 = 0 or 5x + 2 = 0 x = 4 or x = – 2 5 (iii) 2x2 + ax – a2 = 0  2x2 + 2ax – ax – a2 = 0  2x(x + a) – a(x + a) = 0 i.e., (x + a) (2x – a) = 0  x + a = 0 or 2x – a = 0  x = – a or x = Ex.6 Sol.  x=3 a 2 or x=– 4 3 (iii) x(x + 1) + (x + 2) (x + 3) = 42 Solve the following quadratic equations :  x2 + x + x2 + 3x + 2x + 6 – 42 = 0 (i) x2 + 5x = 0  2x2 + 6x – 36 = 0 (ii) x2 = 3x (iii) x2 = 4  x2 + 3x – 18 = 0 (i) x2 + 5x = 0 x(x + 5) = 0  x2 + 6x – 3x – 18 = 0  x = 0 or x + 5 = 0  x(x + 6) – 3(x + 6) = 0  x = 0 or x = – 5  (x + 6) (x – 3) = 0  x = – 6 or x = 3 (ii) x2 = 3x  x2 – 3x = 0 Ex.7  x(x – 3) = 0 Ex.8 Solve for x : 12 abx2 – (9a2 – 8b2) x – 6ab = 0  x = 0 or x = 3 Sol. Given equation is : (iii) x2 = 4 12abx2 – 9a2x + 8b2x – 6ab = 0  x=±2  3ax(4bx – 3a) + 2b(4bx – 3a) = 0 Solve the following quadratic equations :  (4bx – 3a) (3ax + 2b) = 0 (i) 7x2  4bx – 3a = 0 = 8 – 10x (ii) 3(x2 – 4) = 5x (iii) x(x + 1) + (x + 2) (x + 3) = 42 Sol.  x= 3a 4b or 3ax + 2b = 0 or x = – 2b 3a SOLVING A QUADRATIC EQUATION BY COMPLETING THE SQUARE (i) 7x2 = 8 – 10x  7x2 + 10x – 8 = 0 Every quadratic equation can be converted in the form :  7x2 + 14x – 4x – 8 = 0  7x(x + 2) – 4(x + 2) = 0 (x + a)2 – b2 = 0 or (x – a)2 – b2 = 0.  (x + 2) (7x – 4) = 0 Steps :  x+2=0 or 7x – 4 = 0  x=–2 or x= 4 7 (ii) 3(x2 – 4) = 5x 1. Bring, if required, all the term of the quadratic equation to the left hand side. 2. Express the terms containing x as x2 + 2xy or x2 – 2xy.  3x2 – 9x + 4x – 12 = 0 3. Add and subtract y2 to get x2 + 2xy + y2 – y2 or x2 – 2xy + y2 – y2; which gives (x + y)2 – y2 or (x – y)2 – y2.  3x(x – 3) + 4(x – 3) = 0 Thus,  (x – 3) (3x + 4) = 0 (i) x2 + 8x = 0  x2 + 2x × 4 = 0  3x2 – 5x – 12 = 0  x – 3 = 0 or 3x + 4 = 0  x2 + 2x × 4 + 42 – 42 = 0  (x + 4)2 – 16 = 0 (ii) x2 – 8x = 0  x2 – 2 × x × 4 = 0  x2 – 2 × x × 4 + 42 – 42 = 0  (x – 4)2 – 16 = 0 Find the roots of the following quadratic equations (if they exist) by the method of completing the square. (i) 2x2 – 7x + 3 = 0 (i) 2x2 – 7x + 3 = 0  3  3 3  x  – + =0  2  4 4  2  3  i.e.,  x  =0  2   (ii) 4x2 + 4 3 x + 3 = 0  x+ (iii) 2x2 + x + 4 = 0 Sol. i.e., 7 3 x+ =0 2 2  x2 – 3 =0 2  Roots are : [Dividing each term by 2]  x2  x2 – 2 × x × 7  49 3  + =0  x   – 4  16 2   7   49  24   x   –  =0 4   16    7  25  =0 x   – 4  16  2 (ii) 4x2 x– 7 5 =± 4 4 7 5 = = 4 4 or x – 7 5 + 4 4 or x = 7 5 – 4 4 or x = 1 2  x=3 +4 3x+3=0 x2 + i.e., x2 + 2 × x × x +2=0 2 2 1 1 1 +  –  + 2 = 0 4 4 4 2 1 31  =0   x   + 4 16    1  32 31  1  16  2  16  16    2 2  x= x2 + 2 2   3 2 1 1  +2=0  x   – 4 16  2 i.e., x – x=  3  3 and 2 2 2 2 7 7 7 3 +   –  + =0 4 4 4 2 7 25  i.e.,  x   = 4 16   and (iii) 2x2 + x + 4 = 0 7 3 + =0 –2×x× 4 2 2 2 3  3   3  3 + +2×x× – + =0 2  2   2  4 2  EXAMPLES  Ex.9 2 x2 3 3x+ =0 4 7 5 =–  4 4 1 31  i.e.,  x   = – 4 16  This is not possible as the square of a real number can not be negative. SOLVING A QUADRATIC EQUATION BY  USING QUADRATIC FORMULA Hindu Method (Sri Dharacharya Method) : By completing the perfect square as ax2 + bx + c = 0  x2 + b c x+ =0 a a  b  Adding and substracting    2a  2  Ex.11 For a quadratic equation ax2 + bx + c = 0, 2  b  b 2  4ac   x    = 0 2a  4a 2   Which gives, x= where a  0, prove that : x =  b  b 2  4ac 2a ax2 Hence the Quadratic equation (a  0) has two roots, given by = Sol. ax2 + bx + c = 0  4a2x2 + 4abx + 4ac = 0 + bx + c = 0 2 [Multiplying by ‘4a’]  (2ax)2 + 2 × 2ax × b + b2 – b2 + 4ac = 0 2  b  b  4ac  b  b  4ac and  = 2a 2a (2ax + b)2 – b2 + 4ac = 0 (2ax + b)2 = b2 – 4ac Note : Every quadratic equation has at most two and only two real roots.  EXAMPLES  Ex.10 Solve the following quadratic equations by using quadratic formula :  2ax + b = ± b 2  4ac 2ax = – b ± b 2  4ac  b  b 2  4ac 2a x = (i) x2 – 7x + 12 = 0 (ii) 3x2 – x – 10 = 0 Sol. Ex.12 Solve, by using quadratic formula, each of the following equations : (i) Comparing the given equation x2 – 7x + 12 = 0 with standard quadratic equation ax2 + bx + c = 0; we get : a = 1, b = – 7 and c = 12  b2 – 4ac = (–7)2 – 4 × 1 × 12 and  b  b 2  4ac 2a (i) 2x2 + 5 3 x + 6 = 0 (ii) 3x2 + 2 5 x – 5 = 0 Sol. (i) Comparing 2x2 + 5 3 x + 6 = 0 with = 49 – 48 = 1 ax2 + bx + c = 0, we get : b 2  4ac = 1 = 1 a = 2, b = 5 3 and c = 6  x=  b  b 2  4ac 2a  x= 7 1 7 1 7 1 = or = 4 or 3 2 1 2 2 (ii) Comparing the given equation b2 – 4ac = (5 3 )2 – 4 × 2 × 6 = 25 × 3 – 48 = 27 b 2  4ac =  x= 3x2 – x – 10 = 0 with equation ax2 + bx + c = 0; we get : a = 3, b = – 1 and c = – 10  b2 – 4ac = (–1)2 – 4 × 3 × – 10 = 1 + 120 = 121 and b 2  4ac = 121 = 11 Hence, x =  x=  b  b 2  4ac 2a 1  11 1  11 1  11 = or = 6 or –5 2 2 2 3 3 3 = 3 3 27 =  b  b 2  4ac 5 3 3 3 = 2a 24 = 5 3 3 3 4 = 2 3 8 3 3 or =– or – 2 3 4 4 2 or 5 3 3 3 4 (ii) Comparing 3x2 + 2 5 x – 5 = 0 with ax2 + bx + c = 0, we get : a = 3, b = 2 5 and c=–5 b2 – 4ac = ( 2 5 )2 – 4 × 3 × – 5 = 4 × 5 + 60 = 80 b 2  4ac =  x= 80 = 16  5 = 4 5 (i) If b2 – 4ac is a perfect square, the roots are rational and unequal.  b  b 2  4ac 2 54 5 = 2a 23 (ii) If b2 – 4ac is positive but not perfect square, the roots are irrational and unequal. = 2 54 5 6 2 54 5 6 If D = 0, i.e., b2 – 4ac = 0; the roots are real and equal. = 2 5 6 5 5 or = or – 5 6 6 3 If D < 0, i.e., b2 – 4ac < 0 ; i.e., b2 – 4ac is negative; the roots are not real, i.e., the roots are imaginary. or Ex.13 Using the quadratic formula, solve the equation: a2b2x2 – (4b2 – 3a4)x – 12a2b2 = 0 Sol. Ax2 + Bx + C = 0, we get : a2b2, B=– (4b4 – 3a4) (i) 2x2 + 2x + 3 = 0 and C = –12 a2b2 (ii) 2x2 – 7x + 3 = 0  B2 – 4AC = (4b4 – 3a4)2 – 4 × a2b2 × (– 12a2b2) = 16b8 + 9a8 – 24a4b4 + 48a4b4 Sol. (i) Comparing 2x2 + 2x + 3 = 0 B 2  4AC = 4b4 + 3a4 with ax2 + bx + c = 0; we get : a = 2, b = 2 and c=3  B  B 2  4AC 2A D = b2 – 4ac = (2)2 – 4 × 2 × 3 = 4 – 24 = – 20; which is negative. (4b 4  3a 4 )  (4b 2  3a 4 ) = 2  a 2b2 The roots of the given equation are imaginary. (ii) Comparing 2x2 – 7x + 3 = 0 4b 4  3a 4  4b 4  3a 4 = 2a 2 b 2 with ax2 + bx + c = 0; we get : a = 2, b = – 7 and c = 3 4b 4  3a 4  4b 4  3a 4 or 2a 2 b 2 D = b2 – 4ac = (–7)2 – 4 × 2 × 3 = 49 – 24 = 25, which is perfect square. 8b 4  3a 2  6a 4 4b 2 = or or = 2a 2 b 2 a 2 2a 2 b 2 b2 NATURE OR CHARACTER OF THE ROOTS OF A QUADRATIC EQUATION The roots of the given equation are rational and unequal. (iii) Comparing x2 – 5x – 2 = 0  The nature of the roots depends on the value of b2 – 4ac. b2 – 4ac is called the discriminant of the quadratic equation ax2 + bx + c = 0 and is generally, denoted by D.   D= b2 – 4ac (iii) x2 – 5x – 2 = 0 (iv) 4x2 – 4x + 1 = 0 = 16b8 + 9a8 + 24a4b4 = (4b4 + 3a4)2 x =  EXAMPLES  Ex.14 Without solving, examine the nature of roots of the equations : Comparing given equation with A= If D > 0, i..e., b2 – 4ac > 0, i.e., b2 – 4ac is positive; the roots are real and unequal. Also, with ax2 + bx + c = 0; we get : a = 1, b = – 5 and c = – 2 D = b2 – 4ac = (–5)2 – 4 × 1 × – 2 = 25 + 8 = 33 ; which is positive but not a perfect square. The roots of the given equation are irrational and unequal. =  b  b 2  4ac  b  b 2  4ac + 2a 2a =  b  b 2  4ac  b  b 2  4ac 2a = b 2b =– 2a a (iv) Comparing 4x2 – 4x + 1 = 0 with ax2 + bx + c = 0; we get : a = 4, b = – 4, and c = 1 D = b2 – 4ac = (–4)2 – 4 × 4 × 1 = 16 – 16 = 0 And, the product of the roots =  .   Roots are real and equal.   b  b 2  4ac   =    2a   Ex.15 For what value of m, are the roots of the equation (3m + 1) x2 + (11 + m) x + 9 = 0 equal? Sol. Comparing the given equation with ax2 + bx + c = 0; = (b) 2  ( b 2  4ac ) 2 b 2  (b 2  4ac) = 4a 2 4a 2 = b 2  b 2  4ac 4a 2 = 4ac c = a 4a 2 we get : a = 3m + 1, b = 11 + m and c = 9  Discriminant, D = b2 – 4ac = (11 + m)2 – 4(3m + 1) × 9 = 121 + 22m + m2 – 108 m – 36 = m2 – 86m + 85 = m2 – 85m – m + 85  If  and  are the roots of the quadratic equation ax2 + bx + c = 0 ; then : = m(m – 85) – 1 (m – 85) (i) The sum of the roots = (m – 85) (m – 1) =+=– Since the roots are equal, D = 0  (m – 85) (m – 1) = 0 = –  m – 85 = 0 or m – 1 = 0 coefficient of x coefficient of x 2 =  =  SUM AND PRODUCT OF THE ROOTS  Let  and  be the two roots of the quadratic equation ax2 + bx + c = 0. Since, ax2  b  b 2  4ac + bx + c = 0  x = 2a  b  b 2  4ac then, let :  = 2a and  = 2  b  b  4ac 2a  The sum of the roots =  +  b a (ii) The product of the roots  m = 85 or m = 1    b  b 2  4ac      2a   cons tan t term c = a coefficient of x 2 TO CONSTRUCT A QUADRATIC EQUATION WHOSE ROOTS ARE GIVEN x2 – (sum of roots) x + product of roots = 0 To get the quadratic equation with given roots : (i) Find the sum of the roots. (ii) Find the product of the roots. (iii) Substitute the values of steps (i) and (ii) in x2 – (sum of the roots)x + (product of roots) = 0 and get the required quadratic equation.  EXAMPLES   Ex.16 For each quadratic equation given below, find the sum of the roots and the product of the roots : (i) x2 + 3x – 6 = 0 (iii) 3x2 and, the product of the roots = (3) (–3) = – 9  (ii) 2x2 + 5 3 x + 6 = 0  The required quadratic equation is : x2 – (sum of roots) x + (product of roots) = 0 +2 5x–5=0  Sol. (i) Comparing x2 + 3x – 6 = 0  x2 – (0) x + (–9) = 0, i.e., x2 – 9 = 0 (ii) Since, the sum of the roots with ax2 + bx + c = 0, =3+ we get :  The sum of the roots = – = (3 + 3 ) (3 – b 3 =– a 2 And, the product of the roots =  c 6 = =–6 a 1  + 5 3 x + 6 = 0 with ax2 + bx + c = 0 ; we get : b 5 3 =– a 2 2 5  2 5  45 1  = = –  2  2  4 4     The required quadratic equation is : x2 – (sum of roots) x + (product of roots) = 0 with ax2 + bx + c = 0; we get : a = 3, b = 2 5 and c = – 5 b 2 5 =– a 3 and, the product of the roots = c 5 = a 3 Ex.17 Construct the quadratic equation whose roots are given below - (ii) 3 + (iii) 3,3– 3 2 5 2 5 , 2 2 Sol. (i) Since, the sum of the roots = (3) + (–3) = 3 – 3 = 0 2 5 2 5 2 5 2 5 4 + = = =2 2 2 2 2 and, the product of the roots (iii) Comparing 3x2 + 2 5 x – 5 = 0 (i) 3, – 3  x2 – 6x + 6 = 0 = c 6 And, the product of the roots = = =3 a 2 The sum of the roots = –  The required quadratic equation is : (iii) Since, the sum of the roots 3 and c = 6  The sum of the roots = – 3)=9–3=6 x2 – (sum of roots) x + (product of roots) = 0 2x2 a = 2, b = 5 3=6 and, the product of the roots a = 1, b = 3 and c = – 6 (ii) Comparing 3 +3–   1  x2 – 2x +    = 0  4   x2 – 2x – 1 = 0, 4 i.e., 4x2 – 8x – 1 = 0 Ex.18 If a and c are such that the quadratic equation ax2 – 5x + 3 = 0 has 10 as the sum of the roots and also as the product of the roots, find a and c. Sol. For ax2 – 5x + c = 0 the sum of roots = – =– coefficient of x coefficient of x 2 `5 5 = a a and the product of roots = Given : The sum of the roots = 10  b2 – 4ac = 0 constan t term c = 2 a coefficient of x 5 5 1 = 10, i.e., 10a = 5  a = = a 10 2 i.e., (–5)2 – 4 × 3 × p = 0     with ax2 + bx + c = 0; c 1 = 10  c = 10a = 10 × =5 a 2  a= 1 and c = 5 2 As, 2 is one of the roots, x = 2 will satisfy the equation 2x2 + px + 4 = 0  2(2)2 + p(2) + 4 = 0  we get : b2 – 4ac = 0 [Given, that the roots are equal]   64 – 8p2 = 0  – 8p2 = – 64, p2 = 8 and p = ± 8 i.e., p = ± 2 2 Ex.21 If  and  are the roots of the quadratic equation ax2 + bx + c = 0, (a  0) then find the values of :  8 + 2p + 4 = 0 (i) 2 + 2 i.e., 2p = – 12 and p = – 6 (iii) Substituting p = – 6 in the equation 2x2 + px + 4 = 0; we get : 2x2 – 6x + 4 = 0  x2 – 3x + 2 = 0 Sol. . [Dividing each term by 2]  x2 a = 2p, b = – 8 and c = p  (–8)2 – 4 × 2p × p = 0 Ex.19 If one of the roots of the quadratic equation 2x2 + px + 4 = 0 is 2, find the the value of p. also find the value of the other roots. Sol. 25 1 =2 12 12 (ii) Comparing 2px2 – 8x + p = 0 The product of roots = 10   25 – 12p = 0 and p = (ii) 3 + 3   +   We know, sum of roots ( + ) = – And, product of roots () = – 2x – x + 2 = 0 b a c ; therefore : a (i) ( + )2 = 2 + 2 + 2  x(x – 2) (x – 1) = 0  x–2=0 or x–1=0  x=2 or x=1  2 b2 c c b 2  2ac  b =   – 2 = 2 – 2 = a a a a2  a  The other (second) root is 1. Ex.20 In the following, find the value (s) of p so that the given equation has equal roots. (ii) 3 + 3 = ( + ) (2 + 2 – ) (i) 3x2 – 5x + p = 0 (ii) 2px2 – 8x + p = 0 Sol. (i) Comparing 3x2 – 5x + p = 0 with ax2 + bx + c = 0, we get : a = 3, b = – 5 and c = p Since, the roots are equal ; the discriminant  2 + 2 = ( + )2 – 2 (iii)  b =    a  b 2  2ac c      a2 a   b =    a 2  b 2  2ac  ac    = – b(b  3ac)   a2 a3      2  2 + =    b 2  2ac b 2  2ac a b 2  2ac a2 = = × = c c ca a2 a Relation between Roots and Coefficients : If roots of quadratic equation ax2 + bx + c = 0 (a 0) are and  then : (ix) (x) 2 + 2 =  ( + ) 2   (xi)   +       = (i) ( – ) = (  )   4 b 2  4ac  D = a a =± (ii) 2 + 2 = ( + )2 – 2 = (iii) 2 – 2 = ( + ) =– b 2  2ac a2 b b 2  4ac  D = 2 a a 2  4   4 ( 2   2 ) 2  2 2 2 =   2 2  2 2 EQUATIONS REDUCIBLE TO QUADRATIC EQUATIONS  Type 1 : Equations of the form ax4 + bx2 + c = 0, Method : Substitute x2 = y and solve. (  ) 2  4 (iv) 3 + 3 = ( + ) (2 + 2 – ).  EXAMPLES  Ex.22 Solve the following equations : (i) x4 – 26x2 + 25 = 0 (ii) z4 – 10z2 + 9 = 0 Sol. (i) Substituting x2 = y : 3 + 3 = ( + )3 – 3 ( + ) =–    2  2 (  ) 2  2 + = =     x4 – 26x2 + 25 = 0  y2 – 26y + 25 = 0 b(b 2  3ac) a3 i.e., y2 – 25y – y + 25 = 0  y(y – 25) – 1(y – 25) = 0 (v) 3 – 3 = ( – ) (2 + 2 + ). 3 – 3 = ( – )3 + 3 ( – ) i.e., (y – 25) (y – 1) = 0  y – 25 = 0 or y – 1 = 0 = (  ) 2  4 {( + )2 – } i.e., y = 25 or y = 1 2 = 2 (b  ac) b  4ac a3 (vi) 4 + 4 = {( + )2 – 2}2 – 22 2 2  b 2  2ac  c2  –2 =  2  a2  a  4 4 2 2 2 2 (vii)  –  = ( –  ) ( +  ) = (viii) 2 +  b(b 2  2ac) b 2  4ac a4 2 2 + = ( + ) –  y = 25    x2 = 25 | y = 1  x2 = 1  x=±1 x=±5 |  Roots of the given equation are : ± 5, ± 1 (ii) Substituting z2 = x z4 – 10z2 + 9 = 0  x2 – 10x + 9 = 0 i.e., x2 – 9x – x + 9 = 0  x(x – 9) – 1 (x – 9) =0 i.e., (x – 9) (x – 1) = 0  x – 9 = 0 or x – 1 = 0 x = 9  z2 = 9 | x = 1 z2 = 1 z = ± 3 | z = ± 1 Solution of the given equation is : ± 3, ± 1. Type 2 : Equation of the form : px + 1. Square both the sides to get : a – x2 = (bx + c)2 2. Now simplify it to get a quadratic equation. q =r x Method :(i) Multiply each term by x. 3. Solve the quadratic equation obtained. Ex.24 Solve : (ii) Solve the quadratic equation obtained to get the non-zero value(s) of x. Ex.23 Solve :  (ii) 3y + x2 (iii) 2 x 2  9 + x = 13  x + 2x = 1  + 5 = 6x [Multiplying each term by x]  x = 1 + 4x2 – 4x  4x2 – 5x + 1 = 0 i.e., 4x2 – 4x – x + 1 = 0   4x (x – 1) – 1 (x – 1) = 0 i.e., (x – 1) (4x – 1) = 0 x – 1 = 0  (x – 5) (x – 1) = 0 i.e., x – 5 = 0 or 4x – 1 = 0 or x – 1 = 0 x = 5 or x = 1. i.e., x = 1 or x =  Required solution is 5, 1 5 =2 16y   3y × 16y + 5 = 2 × 16y   48y2 – 32y + 5 = 0   48y2 1 4   PROBLEMS ON QUADRATIC EQUATIONS  For solving problems based on quadratic equations, the following steps must be adopted : 1. Read the given statement of the problem carefully to find the required unknown quantity. – 12y – 20y + 5 = 0 i.e., 12y (4y – 1) – 5(4y – 1) = 0  x = 1 – 2x  i.e., x(x – 5) – 1(x – 5) = 0 (ii) 3y + 3x 2  2 + 1 = 2x i.e., 1 + 4x2 – 4x – x = 0  x2 – 6x + 5 = 0 x2 – 5x – x + 5 = 0  (ii) i.e., x = (1 – 2x)2 5 =2 16 y 5 =6 x Sol. (i) x + x + 2x = 1 Sol. (i) 5 =6 x (i) x + (i)  (4y– 1) (12y – 5) = 0 2. Take the unknown quantity as ‘x’ and according to the given statement, form an equation in terms of ‘x’. 3. Simplify and solve the equation to get the value/values of ‘x’. i.e., 4y – 1 = 0 or 12y – 5 = 0 1 5 or y = 4 12   4y = 1 or 12y = 5 i.e., y =  1 5  Required solutions is : , 4 12  EXAMPLES  Ex.25 Find two consecutive natural numbers, whose product is equal to 20. Sol. Type 3 : Let the required two consecutive natural numbers be x and x + 1. Equations involving one radical : Given : x (x + 1) = 20  x2 + x – 20 = 0  (x + 5) (x – 4) = 0  x = – 5, or x = 4 a  x 2 = bx + c Method : Since, x must be a natural number,   x=4  x2 + 32x – 420 = 0 And required numbers are x and x + 1 i.e., and 5. Ex.26 The sum of the squares of two consecutive whole numbers is 61. Find the numbers. Sol. On factorizing, it gives : (x + 42) (x – 10) = 0 i.e., x = – 42 or x = 10 Let the required consecutive whole numbers be x and x + 1.  x = 10  x2 + (x + 1)2 = 61 Hence the larger part = 10 and the smaller part = 16 – x = 16 – 10 = 6 x2 + x2 + 2x + 1 – 61 = 0 2x2 + 2x – 60 = 0  x2 + x – 30 = 0 [Dividing each term by 2]  (x + 6) (x – 5) = 0 [On factorising]  x = – 6, or x = 5 Ex.29 Two positive numbers are in the ratio 2 : 5. If difference between the squares of these numbers is 189 ; find the numbers. Sol. Let numbers be 2x and 5x  x is a whole number,  x = 5   (5x)2 – (2x)2 = 189 And, required numbers are x and x + 1 = 5 and 5 + 1 i.e., 5 and 6   25x2 – 4x2 = 189 and 21x2 = 189 Sol.  8x  x 1 1 8 8 + =  = x 8 x 15 x (8  x ) 15  8 8 = i.e., 120 = 64x – 8x2 2 15 8x  x  x=3 And, required numbers = 2x and 5x = 2 × 3 and 5 × 3 = 6 and 15  x2 – 8x + 15 = 0 [Dividing by 8]  (x – 5) (x – 3) = 0 [On factorizing]  x = 5, or x = 3 When x = 5, the number are x and 8 – x = 5 and 3, and when x = 3, the numbers are x and 8 – x = 3 and 5.  Ex.30 A two digit number is such that the product of the digits is 35. When 18 is added to this number the digits interchange their places. Determine the number. Sol.  8x2 – 64x + 120 = 0  Required numbers are 5 and 3. Ex.28 Divide 16 into two parts such that twice the square of the larger part exceeds the square of the smaller part by 164. Sol. Since, the required numbers are positive, Let the numbers be x and 8 – x.  Let larger part be x, therefore the smaller part = 16 – x Given : 2x2 – (16 – x)2 = 164  2x2 – (256 + x2 – 32x) – 164 = 0 i.e., 2x2 – 256 – x2 + 32x – 164 = 0 189 = 9 x = ± 3 21 i.e., x2 = Ex.27 The sum of two natural numbers is 8. If the 8 , find the two sum of their reciprocals is 15 numbers. Let ten’s digit of the numbers = x and its unit digit = y.   The two digit number is 10x + y. Given : x . y = 35 and 10x + y + 18 = 10y + x   y= 35 and 9x + 18 = 9y x i.e., x + 2 = y On substituting y = x+2= 35 in x + 2 = y ; we get: x 35 x  x2 + 2x = 35 and x2 + 2x – 35 = 0 On factorising, we get : (x + 7)(x – 5) = 0 i.e., x = – 7 or x = 5 Since, x is digit, therefore x = 5 and 35 =7 y = x i.e., 2x2 – 6x = 0 x2 – 3x = 0 i.e., x(x – 3) = 0  The required two digit number = 10x + y x = 0 or x = 3 Since, x = 0 makes the triangle impossible. therefore, x = 3 = 10 × 5 + 7 = 57 Ex.31 The sides (in cm) of a right triangle are x – 1, x and x + 1. Find the sides of triangle. And, sides of the triangle are = x, 2x – 1 and x + 1 = 3, 2 × 3 –1 and 3 + 1 = 3m, 5m Sol. and 4m It is clear that the largest side x + 1 is hypotenuse of the right triangle. Ex.33 If the perimeter of a rectangular plot is 68 m and its diagonal is 26 m. Find its area. x Sol. Let the length of plot = x m D x–1 According to Pythagoras Theorem, we have : x2 + (x – 1)2 = (x + A 1)2   x2 + x2 – 2x + 1 = x2 + 2x + 1  x + breadth =  x(x – 4) = 0 i.e., x = 0 or x = 4 Since, with x = 0 the triangle is not possible; hence x = 4. 68 and breadth = (34 – x) m 2 Given its diagonal = 26 m and we know each angle of the rectangle = 90º.  Sides, are x – 1, x and x + 1 = 4 – 1  x2 + (34 – x)2 = 262 i.e., 3 cm, 4 cm and 5 cm [Applying Pythagoras Theorem] Ex.32 The hypotenuse of a right triangle is 1 m less than twice the shortest side. If the third side is 1 m more than the shortest side, find the sides of the triangle.  x2 + 1156 – 68x + x2 – 676 = 0  2x2 – 68x + 480 = 0  x2 – 34x + 240 = 0 Let the shortest side be x m. i.e., x2 – 34x + 240 = 0 On factorising, we get : (x – 24) (x – 10) = 0 2x + 1 x  2(length + breadth) = perimeter  2(x + breadth) = 68 This gives x2 – 4x = 0 Sol. C (34 – x)m B 26m i.e., x = 24 or x = 10 x = 24   x+1  = (34 – 24) m = 10 m  Hypotensuse = (2x – 1) m and the third side = (x + 1) m Applying Pythagoras theorem, we get ; (2x –  1)2 = x2 + (x + 1)2  4x2 – 4x + 1 = x2 + x2 + 2x + 1  length = 24 m and breadth and, x = 10     length = 10 m and breadth = (34 – 10) m = 24 m Dimensions of the given rectangular plot are 24 m and 10 m. Now, time taken to go 30 km downstream 30 = hrs. 15  x Hence, its area = length × breadth = 24 m × 10m = 240 m2 Ex.34 A train travels a distance of 300 km at a uniform speed. If the speed of the train is increased by 5 km an hour, the journey would have taken two hours less. Find the original speed of the train. Sol. and, time take to come back 30 km upstream 30 = hrs. 15  x Given : the time taken for both the journeys 1 9 = 4 hours 30 min. = 4 hrs = hrs 2 2 Let the original speed of the train be x km/hr. In 1st case, Distance = 300 km and speed = x km/hr.   Time taken =  dis tance 300 = hrs. speed x In 2nd case, Distance = 300 km and speed = (x +5) km/hr.   Time taken = Given :   dis tance 300 hrs. = speed x5  300 300 – =2 x x5   x2 + 5x – 750 = 0 i.e., 450  30x  450  30x 9 = 2 225  x 2  2 × 900 = 9 (225 – x2) i.e., x2 = 225 – 200 x2 = 25 and x = ± 5 Rejecting the negative value of x, we get : x = 5 i.e., the speed of the steam = 5 km/hr i.e., x = – 30 or x = 25 Neglecting x = – 30; we get x = 25 Sol. i.e., x = 25 km per hour Ex.35 A motor boat, whose speed is 15 km/hr in still water, goes 30 km downstream and comes back in a total of 4 hours 30 minutes. Determine the speed of the stream.  30(15  x )  30(15  x ) 9 = (15  x )(15  x ) 2 Ex.36 The hotel bill for a number of people for overnight stay in Rs. 4, 800. If there were 4 people more, the bill each person had to pay would have reduced by Rs. 200. Find the number of people staying overnight. On factorising, we get : (x + 30) (x – 25) = 0 Sol.  2 × 100 = 225 – x2 300x  1500  300x =2 i.e., x 2  5x  2(x2 + 5x) = 1500 30 30 9 + = 15  x 15  x 2 On dividing both the sides by 9, we get : 300( x  5)  300x =2 x ( x  5)   Let the number of people staying overnight be x.   For x people, the hotel bill = Rs 4,800   For 1 person, the hotel bill = Rs 4,800 x When 4 people were more : Let the speed of the stream = x km/hr For (x + 4) people, the hotel bill = Rs 4,800  The speed of the boat downstream = (15 + x) km/hr.  For 1 person, the hotel bill = Rs and, the speed of the boat upstream = (15 – x) km/hr It is given that now the bill paid by each person is reduced by Rs 200.   4,800 4,800 – = 200 x x4 4,800 x4   4,800( x  4)  4,800x = 200 x ( x  4) In 3 years time : The age of son will be (x + 3) years and the age of man will be (3x2 – 12x + 14) + 3 = 3x2 – 12x + 17 years i.e., 200(x2 + 4x) = 4800x + 19200 – 4800x   x2 3x2 – 12x + 17 = 4x + 12 i.e., 3x2 – 16x + 5 = 0 i.e., x2 + 4x – 96 = 0 On factorising, we get : (x – 5) (3x – 1) = 0 On factorising, we get : (x + 12) (x – 8) = 0 i.e., x = 5 or x = i.e., x = – 12 or x=8 1 is not possible ; x = 5 3  No. of people can not be negative Since, x =   No. of people staying overnight = 8 The present age of man = 3x2 – 12x + 14 Sol.  = 3 × 52 – 12 × 5 + 14 = 29 years. And, the present age of son = x = 5 years Ex.39 Find the roots of the equation x2 – 2x – 8 = 0. Sol. Quadratic Equation x2 – 2x – 8 = 0 Let the number of rows be x  After factorization (x – 4) (x + 2) = 0  No. of seats in each rows = x x = 4, – 2  The total number of seats in the auditorium = x × x = x2 Now, the new no. of rows = 2x and the new no. of seats in each row = x – 5 Ex.40 Find the roots of the equation x2 – 4x + 1 = 0. Sol. x= = 2x(x – 5). Given : 2x (x – 5) – x2 = 375 2x2 – 10x – x2 = 375 and x2 – 10x – 375 = 0 On factorising, we get : (x – 25) (x + 15) = 0 i.e., x = 25 or x = – 15 Neglecting x = – 15, we get : no. of rows = 25 Ex.38 Two years ago, a man’s age was three times the square of his son’s age. In three years time, his age will be four times his son’s age. Find their present ages. Let present age of son = x years Two years ago : The age of son was (x – 2) years and so the age of the man was 3(x – 2)2 Here a = 1, b = 4, c = 1 Using Hindu Method The new no. of total seats in the auditorium Sol. 1 3  Ex.37 In an auditorium, the number of rows was equal to the number of seats in each row. If the number of roots is doubled and the number of seats in each row is reduced by5, then the total number of seats is increased by 375. How many rows were there ?  Given : 3x2 – 12x + 17 = 4 (x + 3) 19200 + 4x = = 96 200 4  16  4 =2± 2 3 Ex.41 Find the nature of the roots of the quadratic equation 7x2 – 9x + 2 = 0. Sol. b2 – 4ac = 81– 56 = 25 > 0 and a perfect square so roots are rational and different. Ex.42 Find the nature of the roots of the quadratic equation 2x2 – 7x + 4 = 0. Sol. b2–4ac = 49 – 32 = 17 > 0 (not a perfect square) Its roots are irrational and different. Ex.43 Find the nature of the roots of the quadratic equation x2 – 2 (a + b) x + 2(a2 + b2) = 0. Sol. A = 1, B = –2 (a + b), C = 2 (a2 + b2) B2 – 4AC = 1[2(a + b)]2 – 4(1) (2a2 + 2b2)  Man’s present age = 3 (x – 2)2 + 2 = 4a2 + 4b2 + 8ab – 8a2 – 8b2 = 3(x2 – 4x + 4) + 2 =3x2 – 12x + 14 = – 4a2 – 4b2 + 8 ab  3m – 2m = – 1  m = – 1 = – 4(a–b)2 < 0 So roots are imaginary and different. Ex.48 If  and  are roots of the equation x2 – 5x + 6 = 0 then find the value of 3 + 3. Ex.44 Find the nature of roots of the equation x2 – 2 Sol. Here  +  = 5,  = 6 Sol. 2 x + 1 = 0. The discriminant of the equation Now 3 + 3 = ( + )3 – 3 ( + ) (–2 2 )2 – 4(1) (1) = 8 – 4 = 4 > 0 and a perfect square so roots are real and different but we can't say that roots are rational because coefficients are not rational therefore. = (5)3 – 3.6 (5) = 125 – 90 = 35 ,  = = 2 2 2 2 2 = 2 2 2  2 Ex.49 If the equation (k – 2)x2 – (k – 4) x – 2 = 0 has difference of roots as 3 then find the value of k. ( – ) = Sol. 4 2 (  ) 2  4 Now  +  = ( k  4) 2 ,  = ( k  2) k2  ( – ) = 8 k 4    ( k  2) k 2 2 ±1 2 this is irrational.  the roots are real and different. = Ex.45 Find the nature of the roots of the equation (b + c) x2 – (a + b + c) x + a = 0, (a,b,c  Q) ? Sol. 3 = The discriminant of the equation is (a +b + c)2 – 4(b + c) (a) = a2+ b2+ c2 + 2ab + 2bc + 2ca – 4(b+c)a   ax2 + bx + c = 0 then find the value of So roots are rational and different. Ex.46 If the roots of the equation real then find the value of P. x2  discriminant = Since ,  are the root of the ax2 + bx + c Sol. then a2+ b+ c = 0 – 4 (1) (P) 0 (Since roots are real)  (a+b) = – c /  ...(1) Similarly  P1 Ex.47 If the product of the roots of the quadratic equation mx2 – 2x + (2m–1) = 0 is 3 then find the value of m is Product of the roots c/a = 3 =  (a+b) + c = 0   4 – 4 P  0  4  4P Sol. 1 1 + . (a  b) 2 (a  b) 2 + 2x +P = 0 are Here a = 1, b = 2, c = P (2)2  3k – 6 = ± k Ex.50 If are roots of the equation (a – b – c)2 > 0 Sol. k 2  16  8k  8k  16 ( k  2)  k = 3, 3/2 = a2 + b2 + c2 + 2ab + 2bc + 2ca – 4ab – 4 ac = a2 + b2 + c2 – 2 ab + 2 bc – 2 ca k 2  16  8k  8(k  2) ( k  2) 2m  1 m (a+b) = – c /  ...(2) 1 1 1 1 =   2 2 2 (a  b) (a  b) (c /  ) (c / ) 2  2 2  2  2 (  ) 2  2 + = = c2 c2 c2 c2 = b 2 / a 2  2c / a b 2  2ac = c2 a 2c 2 Ex.51 Find the equation whose roots are 3 and 4. Sol. The quadratic equation is given by Ex.55 If ,  are roots of the equation 2x2 + x –1= 0 then find the equation whose roots are 1/, 1 /  will be Sol.  +  = – 1/2 ,  = – 1/2 x2 – (sum of the roots) x + (product of roots) = 0 The required equation is -  The required equation  1 1 1 =0 x2 –    x +      = x2 – (3 + 4) x + 3.4 = 0 = x2 – 7x + 12 = 0  1  x +  x2 –  =0     Ex.52 Find the quadratic equation with rational coefficients whose one root is 2 + 3 Sol.  1   1  x2 –  2  x + = 0 x2 – x – 2 = 0 1 1     2  2 The required equation is x2 – {(2+ 3 ) + (2– 3)}x + (2 + 3 ) (2– 3)=0 Short cut : Replace x by 1/x or x2 – 4x + 1 = 0  2(1/x)2 + 1/x – 1 = 0  x2 –x–2 = 0 Ex.53 If ,  are the root of a quadratic equation x2 – 3x + 5 = 0 then find the equation whose roots are (2 – 3 + 7) and (2 – 3 + 7) . x2 – 3x + 5 = 0 Sol. So 2 – 3 + 5 = 0 & 2 –3 + 5 = 0 2  SOME MORE EXAMPLES  Ex.56 If r and s are positive, then find the nature of roots of the equation x2 – rx – s = 0 Since ,  are the roots of equation Sol. From the given equation Here Discriminant = r2 + 4s > 0 2  – 3 = – 5 &  – 3= – 5 ( r, s > 0)  roots are real putting in (2 – 3 + 7) & (2 – 3 + 7) ...(1) Again a = 1 > 0 and c = – s < 0 –5 + 7 , – 5 + 7  roots are of opposite signs. 2 and 2 are the roots the required equation is Ex.57 Find the nature of both roots of the equation x2 – 4x + 4 = 0. (x–b) (x–c) + (x–c) (x–a) + (x – a) (x– b) = 0. Ex.54 If , are roots of the equation x2 – 5x + 6 = 0 then find the equation whose roots are  + 3 and  + 3 is Let  + 3 = x Sol. – 5 (x–3) + 6 = 0   x2 – 6 x + 9 –5x + 15 + 6 = 0   x2 –11 x + 30 = 0 3x2 – 2 (a + b + c) x + (ab + bc + ca) = 0 = 4(a + b+ c)2 – 12 (ab + bc + ca) So the required equation is (x – The given equation can be written in the following form : Here discriminant   = x – 3 (Replace x by x – 3) 3)2 Sol. = 4[(a2 + b2 + c2) – (ab + bc + ca)] > 0 ...(1) [ a2 + b2 + c2 > ab + bc+ ca]  Both roots are real. ...(2) Ex.58 If one root of the equation 4x2 + 2x – 1 = 0 is , then find the other root. Sol. Let  and  are roots of the given equation, 1 1 then  +  = – =– –  2 2 Now 42 + 2 – 1 = 0 42 = 1 – 2 ...(1) Now 43 = – 22 = – 1 (1 – 2) 2 [from (1)]  43 – 3 = – 2– =– 1 (1–2) 2 1 –= 2 Ex.59 Find the quadratic equation, whose one root is 1 . 2 5 Sol. Given root = 1 2 5 = 5 –2 So the other root = – 5 – 2. Then sum of the roots = – 4, product of the roots = – 1 Hence the equation is x2 + 4x – 1 = 0