Factoring and Quadratic Equations. Section 1.9 Applications that involve factoring

Chapter 1: Factoring and Quadratic Equations Section 1.9: Applications that involve factoring 1) A number is 20 less than its square. Find all such numbers. Let x = a number then x2 = its square I will replace “a number” with x “is” with an equal sign and “20 less than its square” with x2 - 20 Now I can create an equation. 𝑥 = 𝑥 2 − 20 −𝑥 −𝑥 ________________ 0 = x2 – x – 20 0 = (x + 4)(x – 5) x+4=0 x–5=0 x = -4 x=5 Solution: The numbers are -4 and 5. 3) The square of a number is 6 more than the number. Find all such numbers. A number = x square of a number = x2 “Is” will turn into an equal sign 6 more than a number = x + 6 𝑥2 = 𝑥 + 6 −𝑥 − 6 − 𝑥 − 6 ________________ x2 – x – 6 = 0 (x+2)(x – 3) = 0 x+2=0 x–3=0 x=-2 x=3 Solution: The numbers are -2, 3 Chapter 1: Factoring and Quadratic Equations 5) The product of two consecutive numbers is 72. Find all such numbers. I will call the first of the two numbers x, First number x Since they are consecutive numbers the second number will be called x + 1 Second number x + 1 I will replace the “product of the two numbers” with x(x+1) I will replace the “is” with an equal sign This is the equation I need to solve: x(x+1) = 72 ( I will clear the parenthesis, set equal to 0, then solve by factoring) 𝑥 2 + 𝑥 = 72 −72 − 72 __________________ x2 + x – 72 = 0 (x + 9)(x – 8) = 0 x+9=0 x–8=0 x = -9 x=8 x+1 = -8 x+1=9 Solution: There are two sets of answers. -9 and -8 is one set, and 8 and 9 is the other. Chapter 1: Factoring and Quadratic Equations 7) The product of two consecutive even numbers is 24. Find all such numbers. I will call the first of the two numbers x first number x Since they are consecutive even numbers the second number will be called x + 2 second number x + 2 I will replace the “product of the two numbers” with x(x+2) I will replace the “is” with an equal sign This is the equation I need to solve: x(x+2) = 24 ( I will clear the parenthesis, set equal to 0, then solve by factoring) 𝑥 2 + 2𝑥 = 24 −24 − 24 __________________ x2 + 2x – 24 = 0 (x + 6)(x – 4) = 0 x+6=0 x–4=0 x = -6 x=4 x+2 = -4 x+2=6 Solution: There are two sets of answers. -6 and -4 is one set, and 4 and 6 is the other. Chapter 1: Factoring and Quadratic Equations 9) The product of two consecutive even numbers is 63. Find all such numbers. I will call the first of the two numbers x first number x Since they are consecutive even numbers the second number will be called x + 2 second number x + 2 I will replace the “product of the two numbers” with x(x+2) I will replace the “is” with an equal sign This is the equation I need to solve: x(x+2) = 63 ( I will clear the parenthesis, set equal to 0, then solve by factoring) 𝑥 2 + 2𝑥 = 63 −63 − 63 __________________ x2 + 2x – 63 = 0 (x + 9)(x – 7) = 0 x+9=0 x–7=0 x = -9 x=7 x+2 = -7 x+2=9 Solution: There are two sets of answers. -9 and -7 is one set, and 7 and 9 is the other. Chapter 1: Factoring and Quadratic Equations 11) The length of a rectangular bedroom is 2 feet longer than its width. The area of the bedroom is 120 square feet. Find the dimensions of the room. Width = W Length: L = W + 2 The area is found by multiplying the length and the width This is the equation I need to solve: Area = 120 or LW = 120 (W + 2)(W) = 120 𝑊 2 + 2𝑊 = 120 −120 − 120 _________________ W2 + 2W – 120 = 0 (W + 12)(W – 10) = 0 W + 12 = 0 W – 10 = 0 W = -12 W = 10 The width can’t be (-12). The answer must be the width is 10 feet. Add 2 feet to get the length of 12 feet. Solution: Length 12 feet, Width 10 feet Chapter 1: Factoring and Quadratic Equations 13) A rectangular garden is 4 feet narrower than it is long. The garden has an area of 32 square feet. Find the dimensions of the garden. Length = L Width W = L – 4 Area = LW 32 = L(L-4) 32 = 𝐿2 − 4𝐿 −32 − 32 ______________ 0 = L2 – 4L – 32 0 = (L + 4)(L – 8) L+4=0 L–8=0 L = -4 L=8 The length can’t be (-4). The answer must be the width is 8 feet. Subtract 4 feet to get the width of 4 feet. Solution: Length 8 feet, Width 4 feet Chapter 1: Factoring and Quadratic Equations 15) The base of a triangle is 2 feet longer than its height. The area of the triangle is 7.5 square feet. Find the height of the triangle. Height = h Base = height plus 2 B=h+2 1 Area = 2 ∗ 𝑏𝑎𝑠𝑒 ∗ ℎ𝑒𝑖𝑔ℎ𝑡 1 7.5 = 2 (ℎ + 2)(ℎ) Multiply by 2 to clear the fraction. 1 2*7.5 = 2 ∗ 2 (ℎ + 2)(ℎ) 15 = (h + 2)(h) 15 = ℎ2 + 2ℎ 0 = h2 + 2h – 15 (h + 5)(h – 3) = 0 h+5=0 h–3=0 h = -5 h=3 The height can’t be (-5) so the height must be 3 feet. I am not asked for the base, so I won’t include it in my answer. I could just add 2 to get the base of 5 feet. Solution: height = 3 feet . Chapter 1: Factoring and Quadratic Equations Section 1.9: Applications that involve factoring 17) The height of a triangle is 2 feet shorter than its base. The area of the triangle is 17.5 square feet. Find the height of the triangle. Base = B Height H = B - 2 1 Area = 2 ∗ 𝑏𝑎𝑠𝑒 ∗ ℎ𝑒𝑖𝑔ℎ𝑡 1 17.5 = 2 𝐵(𝐵 − 2) 1 2*17.5 = 2 ∗ 2 𝐵(𝐵 − 2) 35 = B(B – 2) 35 = B2 – 2B 0 = B2 – 2B – 35 0 = (B+5)(B – 7) B+5=0 B–7=0 B = -5 B=7 The base can’t be (-5) so the base must be 7 feet. Subtract 2 to find the height of 5 feet. Solution: Height = 5 feet Chapter 1: Factoring and Quadratic Equations 19) The length of the hypotenuse of a right triangle is 8 inches more than the shortest leg. The length of the longer leg is 7 inches more than the length of the shorter leg. Find the length of each side of the triangle. A = shortest leg B = longer leg C = hypotenuse length of the hypotenuse of a right triangle is 8 inches more than the shortest leg gives C=A+8 length of the longer leg is 7 inches more than the length of the shorter leg gives B=A+7 I will use the A2 + B2 = C2 formula, and replace the B and C with the above values. A2 + (A + 7)2 = (A + 8)2 A2 + (A + 7)(A+7) = (A+8)(A+8) A2 + A2 + 7A + 7A + 49 = A2 + 8A + 8A + 64 2A2 + 14A + 49 = A2 + 16A + 64 -A2 - 16A - 64 -A2 -16A – 64 A2 – 2A – 15 = 0 (A + 3)(A – 5) = 0 A+3=0 A–5=0 A = -3 A=5 The length can’t be (-3), so A must be 5 inches. B = 5 + 7 = 12 inches C = 5 + 8 = 13 inches Solution: short leg 5 inches, longer leg 12 inches, hypotenuse 13 inches. Chapter 1: Factoring and Quadratic Equations 21) The length of a hypotenuse of a right triangle is 1 foot more than the longer leg. The length of the shorter leg is 1 foot less than the length of the longer leg. Find the length of each side of the right triangle. A = shortest leg B = longer leg C = hypotenuse length of a hypotenuse of a right triangle is 1 foot more than the longer leg gives C=B+1 length of the shorter leg is 1 foot less than the length of the longer leg gives A=B-1 I will use the A2 + B2 = C2 formula, and replace the B and C with the above values. (B – 1)2 + B2 = (B + 1)2 (B – 1)(B – 1) + B2 = (B+1)(B+1) B2 – 1B – 1B + 1 + B2 = B2 + 1B + 1B + 1 2B2 - 2B + 1 = B2 + 2B + 1 -B2 - 2B - 1 -B2 – 2B – 1 B2 – 4B = 0 B(B – 4) = 0 B=0 B–4=0 B=4 B can’t be 0 feet, so B must be 4 feet. A = 4 – 1 = 3 feet C = 4 + 1 = 5 feet Solution: short leg 3 feet, longer leg 4 feet, hypotenuse 5 feet Chapter 1: Factoring and Quadratic Equations 23) The length of the hypotenuse in a right triangle is 15 inches. The shortest leg is 3 inches shorter than the length of the longest leg. Find the length of each of the legs. A = shortest leg B - 3 B = longer leg C = 15 (the hypotenuse) (B – 3)2 + B2 = 152 (B – 3)(B – 3) + B2 = 225 B2 – 3B – 3B + 9 + B2 = 225 2B2 – 6B + 9= 225 2B2 – 6B – 216 = 0 This is a trick that will make the work easier. I can divide each number by 2 without changing the solution. B2 - 3B – 108 = 0 This is still hard to factor. The lasts will be -12 and 9. (B– 12)(B+ 9) = 0 B – 12 = 0 B+9=0 B = 12 B = -9 The length can’t be (-9). B must equal 12 inches. A = 12 – 3 = 9 inches Solution: short leg 9 inches, long leg 12 inches, hypotenuse 15 inches. Chapter 1: Factoring and Quadratic Equations 25) The length of the short leg in a right triangle is 3 inches. The longest leg is 1 inch less than the length of the hypotenuse. Find the length of each of the each unknown side. A = 3 inches B=C–1 C=C 32 + (C – 1)2 = C2 9 + (C – 1)(C - 1) = C2 9 + C2 – 1C – 1C + 1 = C2 C2 – 2C + 10 = C2 -C2 - C2 -2C + 10 = 0 +2C +2C 10 = 2C 5 =C I also need to find B: B = C – 1, so B = 4. Solution: short side 3 inches, longer side 4 inches, hypotenuse is 5 inches. Chapter 1: Factoring and Quadratic Equations Chapter 1: Review ONLY WRITE UP ANSWERS, NO NEED TO WORK OUT SOLUTIONS - I DO THEM ALL IN A VIDEO 1) Completely factor the polynomial. State if a polynomial is prime. a) x2 + 7x – 18 b) z2 – 13z + 36 c) -y2 +5y + 14 d) -2y3 – 10y2 + 48y e) 6x2 + 13x + 5 f) x2 + 2x + 3 g) 9n2 + 24n + 15 h) 4x3 – 16x2 – 84x i) 25b2 – 81 j) x2 + 64 k) x3 + 125 l) 27y3 – 64 m) 5m2 + m – 6 n) 3n2 + 7n +2 o) -3x3 – 21x2 + 54x p) 25b2 + 30b + 9 q) x2 + 5x + 2x + 10 r) 5x2 + 10x – 3x – 6 2) Solve each equation. a) 2x(x – 3) = 0 b) (x + 2)(3x – 10)=0 c) x2 – 3x – 10 = 0 d) 5x2 + 16x + 3 =0 e) x(x – 1) = 20 f) 3x(x – 1) – 4 = 14 g) a2 – 49 = 0 h) 25b2 = 16 i) 2x2 + 3x + 1 = 0 j) x(x – 2) = 0 k) y2 + 6y + 9 = 0 l) b2 + 6b = 7 3) The product of two consecutive numbers is 30. Find all such numbers. 4) A rectangular garden is 3 feet narrower than it is long. The garden has an area of 70 square feet. Find the dimensions of the garden. 5) The length of the hypotenuse in a right triangle is 5 inches. The shortest leg is 1 inch shorter than the length of the longest leg. Find the length of each of the legs.