Questions and Answers #16 Equations

Questions and Answers Sheet 16 Equations Question #1 Solve the equation. log(𝑥 + 2) − log 𝑥 = 2 log 4 Answer: Step 1 Given equation log(𝑥 + 2) − log 𝑥 = 2 log 4 Step 2 Formula log 𝑎 log 𝑎 − log 𝑏 = log 𝑏 log(𝑎𝑏) = log 𝑎 × log 𝑏 log 𝑥 𝑛 = 𝑛 log 𝑥 Step 3 We need to solve the equation log(𝑥 + 2) − log 𝑥 = 2 log 4 log(𝑥 + 2) = 2 log 4 log 𝑥 log(𝑥 + 2) = log 42 log 𝑥 Taking anti-log on both sides 𝑥+2 = 16 𝑥 𝑥 + 2 = 16𝑥 16 − 𝑥 = 2 15𝑥 = 2 2 𝑥= 15 Step 4 Answer 2 𝑥= 15 Question #2 Solve the equation. sin 3 𝑥 = 1,0 ≥ 𝑥 < 2 Answer: Step 1 Here , we use the result shown on board. 𝜋 sin 0 = 1 ⇒ 0 = + 2𝑛𝜋, 𝑛 ∈ 𝐼 2 Step 2 For different values of n( where n is integers), we get different solutions of trignometric equation. But only for 𝑛 = 0,1,2, we get solution in [0,2𝜋) sin 3 𝑥 = 1 𝜋 3𝑥 = + 2𝑛𝜋, 𝑛 ∈ 𝐼 2 𝜋 2 𝑥 = + 𝑛𝜋, 𝑛 ∈ 𝐼 6 3 𝜋 𝑛 = 0, 𝑥 = 6 𝜋 2𝜋 5𝜋 𝑛 = 1, 𝑥 = + = 6 3 6 𝜋 4𝜋 6 3 𝑛 = 2, 𝑥 = + = 9𝜋 6 = 3𝜋 2 Step 3 Result: 𝜋 5𝜋 3𝜋 𝑥= , , 6 6 2 Question #3 Solve the rational equation. 5 9 12 − = 2 𝑦 + 4 𝑦 − 4 𝑦 − 16 Answer: Step 1 Given equation 5 9 12 − = 2 𝑦 + 4 𝑦 − 4 𝑦 − 16 Step 2 On solving: 5 9 12 − = 𝑦 + 4 𝑦 − 4 𝑦 2 − 16 5(𝑦 − 4) − 9(𝑦 + 4) 12 ⇒ = 2 (𝑦 + 4)(𝑦 − 4) 𝑦 − 16 5𝑦 − 20 − 9𝑦 − 36 12 [∵ 𝑎2 − 𝑏 2 = (𝑎 − 𝑏)(𝑎 + 𝑏)] ⇒ = 2 (𝑦 2 − 16) 𝑦 − 16 ⇒ 5𝑦 − 20 − 9𝑦 − 36 = 12 ⇒ −4𝑦 = 12 + 36 + 20 ⇒ −4𝑦 = 68 ⇒ 𝑦 = −17 Step 3 Hence, The required value of y is 𝑦 = −17 Question #4 Solve the equation. 9𝑥 − 8 1 + =𝑥 9 𝑥+5 Answer: Step 1 The given equation is, 9𝑥−8 1 + =𝑥 9 𝑥+5 Step 2 First, multiply both sides of the equation by the LCM. 9𝑥 − 8 1 ∗ 9(𝑥 + 5) + ∗ 9(𝑥 + 5) = 𝑥 ∗ 9(𝑥 + 5) 9 𝑥+5 (9𝑥 − 8) ∗ (𝑥 + 5) + 9 = 9𝑥(𝑥 + 5) (9𝑥 − 8) ∗ (𝑥 + 5) + 9 = 9𝑥 2 + 45𝑥 9𝑥 2 + 37𝑥 − 31 = 9𝑥 2 + 45𝑥 Step 3 Now, simplify the obtained equation as follows. 9𝑥 2 + 37𝑥 − 31 + 31 = 9𝑥 2 + 45𝑥 + 31 9𝑥 2 + 37𝑥 = 9𝑥 2 + 45𝑥 + 31 −8𝑥 = 31 𝑥= −31 8 Question #5 Solve the rational equation. 2𝑥 𝑥+1 − 3 𝑥+5 = −8𝑥 2 𝑥 2 +6𝑥+5 Answer: 2𝑥 𝑥+1 − 3 𝑥+5 = −8𝑥 2 𝑥 2 +6𝑥+5 Solving the given expression: 2𝑥 3 −8𝑥 2 ⇒ − = 2 𝑥 + 1 𝑥 + 5 𝑥 + 6𝑥 + 5 2𝑥(𝑥 + 5) − 3(𝑥 + 1) −8𝑥 2 ⇒ = 2 (𝑥 + 1)(𝑥 + 5) 𝑥 + 6𝑥 + 5 2𝑥 2 + 10𝑥 − 3𝑥 − 3 −8𝑥 2 ⇒ = 𝑥 2 + 5𝑥 + 𝑥 + 5 𝑥 2 + 6𝑥 + 5 2 2𝑥 + 7𝑥 − 3 −8𝑥 2 ⇒ 2 = 2 𝑥 + 6𝑥 + 5 𝑥 + 6𝑥 + 5 ⇒ 2𝑥 2 + 7𝑥 − 3 = −8𝑥 2 ⇒ 2𝑥 2 + 8𝑥 2 + 7𝑥 − 3 = 0 ⇒ 10𝑥 2 + 7𝑥 − 3 = 0 ⇒ 10𝑥 2 + 10𝑥 − 3𝑥 − 3 = 0 ⇒ 10𝑥(𝑥 + 1) − 3(𝑥 + 1) = 0 ⇒ (10𝑥 − 3)(𝑥 + 1) = 0 3 ⇒ 𝑥 = −1, 10 Question #6 Solve the rational equation. 𝑥 𝑥+1 − 3 𝑥+3 = −2𝑥 2 𝑥 2 +4𝑥+3 Answer: 𝑥 𝑥+1 − 3 𝑥+3 = −2𝑥 2 𝑥 2 +4𝑥+3 Solving the given expression: 𝑥 3 −2𝑥 2 ⇒ − = 2 𝑥 + 1 𝑥 + 3 𝑥 + 4𝑥 + 3 𝑥(𝑥 + 3) − 3(𝑥 + 1) −2𝑥 2 ⇒ = 2 (𝑥 + 1)(𝑥 + 3) 𝑥 + 4𝑥 + 3 𝑥 2 + 3𝑥 − 3𝑥 − 3 −2𝑥 2 ⇒ 2 = 2 𝑥 + 3𝑥 + 𝑥 + 3 𝑥 + 4𝑥 + 3 𝑥2 − 3 −2𝑥 2 ⇒ 2 = 𝑥 + 4𝑥 + 3 𝑥 2 + 4𝑥 + 3 2 ⇒ 𝑥 − 3 = −2𝑥 2 ⇒ 𝑥 2 + 2𝑥 2 − 3 = 0 ⇒ 3𝑥 2 − 3 = 0 ⇒ 3(𝑥 2 − 1) = 0 ⇒ (𝑥 + 1)(𝑥 − 1) = 0 ⇒ 𝑥 = −1,1 Question #7 Solve the differential equation. 𝑦 ′′ + (𝑦 ′ )2 = 2𝑒 −𝑦 Answer: Step 1 Given: 𝑦 ′′ + (𝑦 ′ )2 = 2𝑒 −𝑦 Step 2 On considering the given differential equation as follows, 𝑦 ′′ + (𝑦 ′ )2 = 2𝑒 −𝑦 ...(1) Suppose 𝑢 = 𝑒 𝑦 Then, 𝑦 = log 𝑢 1 𝑦 ′ = ∗ 𝑢′ 𝑢 𝑢𝑢′′ −𝑢′∗𝑢′ 𝑦 ′′ = 𝑣 𝑢 Step 3 Equation (1) becomes as follows, 𝑣 𝑢𝑢′′ − (𝑢′ )𝑣 𝑢′ 2 + ( ) = 𝑣 𝑢 𝑢 𝑢 𝑢′′ = 2 On integrating twice 𝑢′ = 2𝑥 + 𝑐1 𝑢 = 𝑥 2 + 4𝑥 + 𝑐2 𝑒 𝑦 = 𝑥 2 + 𝑐1 𝑥 + 𝑐2 Question #8 Solve the equation. 𝑥 2 + 7𝑥 + 10 = 0 Answer: Step 1 Given equation is 𝑥 2 + 7𝑥 + 10 = 0 . To solve the given equation. Solution: Solving the given equation. 𝑥 2 + 7𝑥 + 10 = 0 𝑥 2 + 5𝑥 + 2𝑥 + 10 = 0 𝑥(𝑥 + 5) + 2(𝑥 + 5) = 0 (𝑥 + 5)(𝑥 + 2) = 0 (𝑥 + 5) = 0 or (𝑥 + 2) = 0 𝑥 = −5 or 𝑥 = −2 Therefore, solution is 𝑥 = −5 or 𝑥 = −2. Step 2 Hence, required solution is 𝑥 = −5, −2. Question #9 Solve the equation. 1 𝑥 1 ( ) = 4 64 Answer: Step 1 1 𝑥 1 4 64 To solve the equation: ( ) = Solution: Given equation is, 1 𝑥 1 4 64 ( ) = On simplifying further we get: 1 𝑥 1 4 64 ( ) = 1 𝑥 1 ⇒( ) = 3 4 4 1 𝑥 1 3 1 4 4 2 ⇒ ( ) = ( ) As base ( ) is equal on both sides of the equation,we can equate powers ⇒𝑥=3 Step 2 Result: Required solution is: 𝑥 = 3. Question #10 Solve the radical equation. √𝑥 + 16 − √𝑥 − 4 = 2 Answer: Step 1 Solve the radical √𝑥 + 16 − √𝑥 − 4 = 2 Step 2 √𝑥 + 16 − √𝑥 − 4 = 2 ⇒ √𝑥 + 16 = √𝑥 − 4 + 2 squaring both sides 2 2 (√𝑥 + 16) = (√𝑥 − 4 + 2) Expanding both sides 𝑥 + 16 = 𝑥 − 4 + 2(√𝑥 − 4)(2) + 4 [(𝑎 + 𝑏)2 = 𝑎2 + 2𝑎𝑏 + 𝑏 2 ] ⇒ 𝑥 + 16 = 𝑥 + 4√𝑥 − 4 ⇒ 16 = 4√𝑥 − 4 16 ⇒ = √𝑥 − 4 4 ⇒ 4 = √𝑥 − 4 squaring both sides 16 = 𝑥 − 4 ⇒ 20 = 𝑥 Question #11 Solve the rational equation. 𝑥 10 −6= 𝑥 − 10 𝑥 − 10 Answer: Step 1 Consider the equation 𝑥 𝑥−10 −6= 10 𝑥−10 . multiply both sides by (𝑥 − 10) and simplify it 𝑥 10 ) (𝑥 − 10) − 6(𝑥 − 10) = ( ) (𝑥 − 10) 𝑥 − 10 𝑥 − 10 𝑥 − 6(𝑥 − 10) = 10 Step 2 use distributive property on −6(𝑥 − 10) and combine the like term 𝑥 − 6𝑥 + 60 = 10 −5𝑥 + 60 = 10 subtract 60 both sides and simplify it −5𝑥 + 60 − 60 = 10 − 60 −5𝑥 = −50 Now divide both sides by -5 and simplify it −5𝑥 −50 = ( −5 −5 𝑥 = 10 Hence, the value of 𝑥 = 10. Question #12 Solve the rational equation. 1 90 𝑥 𝑥2 1+ = Answer: Step 1 Given, 1 90 1+ = 2 𝑥 𝑥 Step 2 On simplification, we get 𝑥+1 90 ⇒ = 2 [by taking LCM] 𝑥 𝑥 90 ⇒𝑥+1 = 𝑥 ⇒ 𝑥 2 + 𝑥 = 90 ⇒ 𝑥 2 + 𝑥 − 90 = 0 Step 3 Now solving the obtained quadratic equation: ⇒ 𝑥 2 + 10𝑥 − 9𝑥 − 90 = 0 ⇒ 𝑥(𝑥 + 10) − 9(𝑥 + 10) = 0 ⇒ (𝑥 + 10)(𝑥 − 9) = 0 ⇒ 𝑥 + 10 = 0 or 𝑥 − 9 = 0 ⇒ 𝑥 = −10 or 𝑥 = 9 Question #13 Solve the rational equation. 7 𝑥 +𝑥 = 88 𝑥 Answer: Step 1 In first step, simplify the equation to form a more specific form where variables are on one side and constant values on the other side of equality as. 7 88 +𝑥 = 𝑥 𝑥 multiply by x on both sides of equation; 7 88 𝑥 𝑥 ( + 𝑥) 𝑥 = ( ) 𝑥 7 + 𝑥 2 = 88 𝑥 2 = 88 − 7 𝑥 2 = 81 Step 2 Now solve for x, to get the desired values as. 𝑥 2 = 81 𝑥 = ±√81 𝑥 = ±9 𝑥 = 9, −9 Question #14 Solve the rational equation. 𝑥 𝑥 − =5 4 6 Answer: Step 1 Here the given equation is 𝑥/4 – 𝑥/6 = 5 First find the common denominator, so the LCM of the denominator 4 and 6 is 12. So: 𝑥 𝑥 − =5 4 6 Step 2 Multiply by LCM 12 𝑥 𝑥 ∗ 12 − ∗ 12 = 5 ∗ 12 4 6 3𝑥 − 2𝑥 = 60 𝑥 = 60 Step 3 Hence , the value of x is 60. Question #15 Solve the differential equation. 𝑑𝑦 5√𝑥𝑦 = 1, 𝑥, 𝑦 > 0 𝑑𝑥 Answer: Step 1 Given, 5√𝑥𝑦 𝑑𝑦 = 1, 𝑥, 𝑦 > 0 𝑑𝑥 Step 2 Rewriting the equation, 1 5√𝑦𝑑𝑦 = 𝑑𝑥 √𝑥 Step 3 Integrate both sides, 1 ⇒ 5 ∫ √𝑦𝑑𝑦 = ∫ 𝑑𝑥 + 𝐶, √𝑥 1 1 𝑦 2+1 𝑥 −2+1 ⇒ 5( )=( )+𝐶 1 1 +1 − +1 2 2 3 1 2𝑦 2 ⇒ 5( ) = 2𝑥 2 + 𝐶 3 3 ⇒ 𝑦2 = 1 3 (2𝑥 2 + 𝐶) 10 𝐶 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡. 3 ⇒ 𝑦 = ( (2𝑥 10 1 +𝐶 2 2 3 )) Question #16 Solve the following equation. 4 =3 √𝑥 − 8 − √𝑥−8 Answer: Step 1 4 Given, √𝑥 − 8 − =3 √𝑥−8 Step 2 2 (√𝑥−8) −4 √𝑥−8 =3 Step 3 𝑥−8−4 =3 √𝑥−8 Step 4 𝑥 − 12 = 3√𝑥 − 8 Step 5 Squaring on both sides, we get Step 6 2 (𝑥 − 12)2 = (3√𝑥 − 8) Step 7 𝑥 2 − 24𝑥 + 144 = 9(𝑥 − 8) Step 8 𝑥 2 − 24𝑥 + 144 = 9𝑥 − 72 Step 9 𝑥 2 − 24𝑥 − 9𝑥 + 144 + 72 = 0 𝑥 2 − 33𝑥 + 216 = 0 Step 10 𝑥 2 − 9𝑥 − 24𝑥 + 216 = 0 𝑥(𝑥 − 9) − 24(𝑥 − 9) = 0 (𝑥 − 9)(𝑥 − 24) = 0 𝑥 − 9 = 0 𝑜𝑟 𝑥 − 24 = 0 𝑥 = 9 𝑜𝑟 𝑥 = 24 Question #17 Find all real solutions of the equation. 𝑥 5 + 8𝑥 2 = 0 Answer: Step 1 To find all the real solution of the equation: 𝑥 5 + 8𝑥 2 = 0 Solution: Given equation is 𝑥 5 + 8𝑥 2 = 0 . Solving the given equation. 𝑥 5 + 8𝑥 2 = 0 𝑥 2 (𝑥 3 + 8) = 0 𝑥 2 = 0 or 𝑥 3 + 8 = 0 𝑥 = 0 or 𝑥 3 = −8 𝑥 = 0 or 𝑥 = −2 Therefore, solution of the given equation are 𝑥 = −2,0. Step 2 Hence, required solution is 𝑥 = −2,0. Question #18 Solve the equation. 𝑒 4𝑥 + 4𝑒 2𝑥 − 21 = 0 Answer: Step 1 Let us suppose 𝑦 = 𝑒 2𝑥 , substitute the value of y in the given equation: 𝑦 = 𝑒 2𝑥 𝑒 4𝑥 + 4𝑒 2𝑥 − 21 = 0 (𝑒 2𝑥 )2 + 4(𝑒 2𝑥 ) − 21 = 0 𝑦 2 + 4𝑦 − 21 = 0 Step 2 Now factorising the equation 𝑦 2 + 4𝑦 − 21 = 0 . 𝑦 2 + 4𝑦 − 21 = 0 (𝑦 + 7)(𝑦 − 3) = 0 Step 3 Substituting 𝑒 2𝑥 back in the obtained equation (𝑦 + 7)(𝑦– 3) = 0 (𝑦 + 7)(𝑦 − 3) = 0 (𝑒 2𝑥 + 7)(𝑒 2𝑥 − 3) = 0 ⇒ (𝑒 2𝑥 + 7) = 0 �𝑛𝑑 (𝑒 2𝑥 − 3) = 0 (𝑒 2𝑥 + 7) = 0 ⇒ 𝑒 2𝑥 = −7 2𝑥 = ln(−7) 𝑥= ln(−7) 2 ; not be a solution. Similarily: (𝑒 2𝑥 − 3) = 0 𝑒 2𝑥 = 3 3𝑥 = ln(3) 𝑥= ln(3) 3 Step 4 A negative argument in logarithm will not give a real solution. Therefore, the solution of the given equation is: 𝑥= ln(3) 3 Question #19 Solve the equation. 𝑥 2 − 14𝑥 = 0 Answer: Step 1 Given equation is 𝑥 2 − 14𝑥 = 0 . To solve the given equation. Solution: Solving the given equation. 𝑥 2 − 14𝑥 = 0 𝑥(𝑥 − 14) = 0 𝑥 = 0 or (𝑥 − 14) = 0 𝑥 = 0 or 𝑥 = 14 Therefore, solution is 𝑥 = 0,14. Step 2 Hence, required solution is 𝑥 = 0,14. Question #20 Describe in words the surface whose equation is given. 0 = 𝜋/4 Answer: Consider the equation we are given. Since angle is fixed, it means that only two dimensions can vary. These are r and z. This means that the surface is a vertical half plane since 𝑟 > 0 and z can be anything. Result: The describer surface is a half-plane.