Questions and Answers #5 Quadratics

Questions and Answers Sheet 5 Quadratics Question #1 To solve: 𝑥(5𝑥 + 2) = 3 Answer: Given: 𝑥(5𝑥 + 2) = 3 By distributive property, 𝑥(5𝑥 + 2) = 𝑥 × 5𝑥 + 𝑥 × 2 𝑥(5𝑥 + 2) = 5𝑥 2 + 2𝑥 Replace x(5x+2) with 5𝑥 2 + 2𝑥 To solve 5𝑥 2 + 2𝑥 = 3 Subtract 3 on both sides, 5𝑥 2 + 2𝑥 − 3 = 3 − 3 5𝑥 2 + 2𝑥 − 3 = 0 To solve, 5𝑥 2 + 2𝑥 − 3 = 0 Split the middle term +2𝑥 as + 5𝑥 and − 3𝑥 Replace +2𝑥 with + 5𝑥 − 3𝑥 in 5𝑥 2 + 2𝑥 − 3 = 0 5𝑥 2 + 5𝑥 − 3𝑥 − 3 = 0 The above equation can be written as 5𝑥 × 𝑥 + 1 × 5𝑥 − 3 × 𝑥 − 3 × 1 Apply distributive property for first two terms and last two terms, 5𝑥 × 𝑥 + 1 × 5𝑥 − 3 × 𝑥 − 3 × 1 = 5𝑥(𝑥 + 1) − 3(𝑥 + 1) Again apply distributive property for 5x(x+1)-3(x+1) 5𝑥(𝑥 + 1) − 3(𝑥 + 1) = (5𝑥 − 3)(𝑥 + 1) So the factors are (5x-3) and (x+1) 5𝑥 2 + 2𝑥 − 3 = (5𝑥 − 3)(𝑥 + 1) To find the value of x, Take x+1=0 Subtract 1 on both sides, x+1-1=0-1 x=-1 Take 5x-3=0 Add 3 on both sides, 5x-3+3=0+3 5x=3 5𝑥 3 Divide by 5 on both sides, = 5 5 3 𝑥= 5 3 So, the values of x are and -1 5 Question #2 To solve: 10𝑥 − 2(𝑥 + 4) = 8(𝑥 − 2) + 6 Answer: Given: 10𝑥 − 2(𝑥 + 4) = 8(𝑥 − 2) + 6 By distributive property, −2(𝑥 + 4) = −2 × 𝑥 + (−2) × 4 −2(𝑥 + 4) = −2𝑥 − 8 By distributive property, 8(𝑥 − 2) = 8 × 𝑥 − 8 × 2 8(𝑥 − 2) = 8𝑥 − 16 Replace -2(x+4) with -2x-8 Replace 8(x-2) with 8x-16 10𝑥 − 2(𝑥 + 4) = 10 − 2𝑥 − 8 10𝑥 − 2(𝑥 + 4) = 8𝑥 − 8 8(𝑥 − 2) + 6 = 8𝑥 − 16 + 6 8(𝑥 − 2) + 6 = 8𝑥 − 10 8𝑥 − 8 = 8𝑥 − 10 Add -8x+8 on both sides, 8𝑥 − 8 − 8𝑥 + 8 = 8𝑥 − 10 − 8𝑥 + 8 Add the like terms, 0x+0=-2+0x 0𝑥 = −2 As there is no x term, no solution exists for the equation. Question #3 To solve: 3 8 𝑏 5 3 12 + = Answer: Multiply by 2 on both sides, 3 𝑏 5 24 ( + ) = × 24 8 3 12 By distributive property, 3 𝑏 3 𝑏 8 3 3 𝑏 8 3 8 3 3 𝑏 8 3 24 ( + ) = 24 + 24 24 ( + ) = 3 × 3 + 𝑏 × 8 24 ( + ) = 9 + 8𝑏 5 12 5 12 × 24 = 5 × 2 × 24 = 10 The value of 5 12 3 × 24 is 10 𝑏 Replace 24 ( + ) with 9+8b 8 Replace 5 12 3 × 24 with 10 9+8b=10 Subtract 9 on both sides, 9+8b-9=10-9 Add the like terms, 8b+0=1 8b=1 Divide by 8 on both sides, So, the value of b is 1 8𝑏 8 = 8 Question #4 To solve: 𝑥 2 − 6𝑥 = 𝑥(8 + 𝑥) 1 8 Answer: Calculation: By distributive property, 𝑥(8 + 𝑥) = 𝑥 ⋅ 8 + 𝑥 ⋅ 𝑥 𝑥(8 + 𝑥) = 8𝑥 + 𝑥 2 Replace 𝑥(8 + 𝑥) with 8𝑥 + 𝑥 2 𝑥 2 − 6𝑥 = 8𝑥 + 𝑥 2 Add −𝑥 2 − 8𝑥 on both sides, 𝑥 2 − 6𝑥 − 𝑥 2 − 8𝑥 = 8𝑥 + 𝑥 2 − 𝑥 2 − 8𝑥 Add the like terms, 0𝑥 2 − 14𝑥 = 0𝑥 2 + 0𝑥 -14x=0 x=0 The value of x is 0. Question #5 To solve: 𝑥2 2 + 𝑥 20 = 1 10 Answer: Calculation: Multiply by 20 on both sides, 20 ( 𝑥2 2 + 𝑥 20 )= 1 10 ⋅ 20 By distributive property, 𝑥2 𝑥 𝑥2 𝑥 20 ( + ) = 20 ⋅ + 20 ⋅ 2 20 2 20 20 ( 1 10 𝑥2 2 + 𝑥 20 ) = 10𝑥 2 + 𝑥 ⋅ 20 = 2 10𝑥 2 + 𝑥 = 2 Subtract 2 on both sides, 10𝑥 2 + 𝑥 − 2 = 2 − 2 10𝑥 2 + 𝑥 − 2 = 0 Split the middle term +𝑥 as + 5𝑥 and − 4𝑥. Replace +𝑥 with − 4𝑥 + 5𝑥 in 10𝑥 2 + 𝑥 − 2 = 0 10𝑥 2 − 4𝑥 + 5𝑥 − 2 = 0 The above equation can be written as 2𝑥 ⋅ 5𝑥 − 2 ⋅ 2𝑥 + 1 ⋅ 5𝑥 − 1 ⋅ 2 Apply distributive property for first two terms and last two terms, 2𝑥 ⋅ 5𝑥 − 2 ⋅ 2𝑥 + 1 ⋅ 5𝑥 − 1 ⋅ 2 = 2𝑥(5𝑥 − 2) + 1(5𝑥 − 2) Again apply distributive property for 2𝑥(5𝑥 − 2) + 1(5𝑥 − 2) 2𝑥(5𝑥 − 2) + 1(5𝑥 − 2) = (2𝑥 + 1)(5𝑥 − 2) So the factors are (2𝑥 + 1) and (5𝑥 − 2) 10𝑥 2 + 𝑥 − 2 = (2𝑥 + 1)(5𝑥 − 2) To find the value of x, Take 5x-2=0 Add 2 on both sides, 5x-2+2=0+2 Add the like terms, 5x=2 Divide by 5 on both sides, 5𝑥 5 = 𝑥= 2 5 2 5 Take 2x+1=0 Subtract 1 on both sides, 2x+1-1=0-1 Add the like terms, 2x=-1 2𝑥 1 Divide by 2 on both sides, = − 2 2 1 𝑥=− 2 1 2 The factors are 𝑥 = − and𝑥 = . 2 1 5 2 2 5 So, the values of x are − and . Question #6 To solve: 𝑦2 = 30 𝑦 15 + 1 2 Answer: Calculation: Multiply by 30 on both sides, 30 ⋅ 𝑦2 30 =( 𝑦 15 1 + ) 30 2 By distributive property, ( ( 𝑦 15 𝑦 15 1 𝑦 2 1 15 + ) 30 = 1 ⋅ 30 + ⋅ 30 2 + ) 30 = 2𝑦 + 15 30 ⋅ 2 𝑦2 30 = 𝑦2 𝑦 2 = 2𝑦 + 15 Subtract -2y-15 on both sides, 𝑦 2 − 2𝑦 − 15 = 2𝑦 + 15 − 2𝑦 − 15 𝑦 2 − 2𝑦 − 15 = 0 Split the middle term −2𝑦 as − 5𝑦 and + 3𝑦 Replace -2y with −5𝑦 + 3𝑦 in 𝑦 2 − 2𝑦 − 15 = 0, 𝑦 2 − 5𝑦 + 3𝑦 − 15 = 0 The above equation can be written as 𝑦 ⋅ 𝑦 − 5 ⋅ 𝑦 + 3 ⋅ 3 − 3 ⋅ 5 Apply distributive property for first two terms and last two terms, 𝑦 ⋅ 𝑦 − 5 ⋅ 𝑦 + 3 ⋅ 𝑦 − 3 ⋅ 5 = 𝑦(𝑦 − 5) + 3(𝑦 − 5) Again apply distributive property for 𝑦(𝑦 − 5) + 3(𝑦 − 5) 𝑦(𝑦 − 5) + 3(𝑦 − 5) = (𝑦 + 3)(𝑦 − 5) So the factors are (𝑦 + 3) and (𝑦 − 5) 𝑦 2 − 2𝑦 − 15 = (𝑦 + 3)(𝑦 − 5) To find the value of y, Take y-5=0 Add 5 on both sides, y-5+5=0+5 Add the like terms, y=5 Take y+3=0 Subtract 3 on both sides, y+3-3=0-3 Add the like terms, y=-3 The factors are 𝑦 = −3 and 𝑦 = 5 So, the values of y are -3 and 5. Question #7 To solve: 4𝑡 2 5 𝑡 3 5 10 = + Answer: Step 1 4𝑡 2 Given equation: 5 𝑡 3 5 10 = + Multiply by 10 on both sides, 4𝑡 2 10 × 5 𝑡 3 5 10 =( + ) 10 By distributive property, 𝑡 3 5 𝑡 10 3 5 10 4𝑡 2 ( + ( + 10 × 𝑡 3 5 10 ) 10 = 10 × + 10 × ) 10 = 2𝑡 + 3 5 = 8𝑡 2 8𝑡 2 = 2𝑡 + 3 Subtract -2t-3 on both sides, 8𝑡 2 − 2𝑡 − 3 = 2𝑡 + 3 − 2𝑡 − 3 8𝑡 2 − 2𝑡 − 3 = 0 Split the middle term -2t as -6t and +4t Replace -2t with -6t+4t in 8𝑡 2 − 2𝑡 − 3 = 0 8𝑡 2 − 6𝑡 + 4𝑡 − 3 = 0 Step 2 The above equation can be written as 2𝑡 × 4𝑡 − 2𝑡 × 3 + 1 × 4𝑡 − 3 × 1 Apply distributive property for first two terms and last two terms, 2𝑡 × 4𝑡 − 2𝑡 × 3 + 1 × 4𝑡 − 3 × 1 = 2𝑡(4𝑡 − 3) + 1(4𝑡 − 3) Again apply distributive property for 2𝑡(4𝑡 − 3) + 1(4𝑡 − 3) 2𝑡(4𝑡 − 3) + 1(4𝑡 − 3) = (2𝑡 + 1)(4𝑡 − 3) So the factors are (2t+1) and (4t-3) 8𝑡 2 − 2𝑡 − 3 = (2𝑡 + 1)(4𝑡 − 3) To find the value of t, Take 4t-3=0 Add 3 on both sides, 4t-3+3=0+3 Add the like terms, 4t=3 Divide by 4 on both sides, 4𝑡 3 = 4 𝑡= 4 3 4 Take 2t+1=0 Subtract 1 on both sides, 2t+1-1=0-1 Add the like terms, 2t=-1 2𝑡 1 Divide by 2 on both sides, = − 2 1 The factors are 𝑡 = − and 𝑡 = 2 So, the values of t are − 1 2 and 2 3 4 3 4 Question #8 To calculate: The break-even point for 𝑃(𝑥) = −1100 + 120𝑥 − 𝑥 2 Answer: Step 1 Given Information: Profit function, 𝑃(𝑥) = −1100 + 120𝑥 − 𝑥 2 Formula Used: Break-even point is the point, when profit is zero or revenue is equal to the cost. Profit = Total revenue − Total cost Zero product property: The General form of quadratic equation is 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 Where a, b, c are constants and 𝑎 ≠ 0 For real numbers a and b. ab=0 if and only if a=0 or b=0 or both. Step 2 Calculation: To find the break-even point, put revenue is equal to the cost. P(x)=0 −1100 + 120𝑥 − 𝑥 2 = 0 𝑥 2 − 110𝑥 − 10𝑥 + 1100 = 0 Take the common terms outside the parenthesis 𝑥(𝑥 − 110) − 10(𝑥 − 110) = 0 (𝑥 − 10)(𝑥 − 110) = 0 Use the zero product property, x-10=0 or x-110=0 x=10 or x=110 It is given that 𝑥 ≤ 100 Hence, the break-even point is 10 units. Question #9 To solve: 𝑥 2 + 11𝑥 + 24 = 0 Answer: Given: 𝑥 2 + 11𝑥 + 24 = 0 Split the middle term 11x as +3x and +8x Replace 11x with +3x+8x in 𝑥 2 + 11𝑥 + 24 = 0 𝑥 2 + 8𝑥 + 3𝑥 + 24 = 0 The above equation can be written ads 𝑥 × 𝑥 + 8 × 𝑥 + 3 × 𝑥 + 3 × 8 Apply distributive property for first two terms and last two terms, 𝑥 × 𝑥 + 8 × 𝑥 + 3 × 𝑥 + 3 × 8 = 𝑥(𝑥 + 8) + 3(𝑥 + 8) Again apply distributive property for 𝑥(𝑥 + 8) + 3(𝑥 + 8) 𝑥(𝑥 + 8) + 3(𝑥 + 8) = (𝑥 + 3)(𝑥 + 8) So the factors are (x+8) and (x+3) 𝑥 2 + 11𝑥 + 24 = (𝑥 + 8)(𝑥 + 3) To find the value of x Take x+8=0 Subtract 8 on both sides, x+8-8=0-8 x=-8 Take x+3=0 Subtract 3 on both sides, x-3+3=0-3 x=-3 So, the values of x are -8 and -3 Question #10 To determine: 13𝑥 − 15𝑥 + 8 = 4𝑥 + 2 − 24 Answer: To solve 13𝑥 − 15𝑥 + 8 = 4𝑥 + 2 − 24 +8 − 2𝑥 = 4𝑥 − 22 Add 2x+22 on both sides, +8 − 2𝑥 + 2𝑥 + 22 = 4𝑥 − 22 + 2𝑥 + 22 Add the like terms, 30 + 0𝑥 = 6𝑥 + 0 30 = 6𝑥 So, it is 6x=30 Divide by 6 on both sides, 6𝑥 30 = 6 𝑥= 6 30 6 Divide by 6 on both numerator and denominator, 30 6 𝑥= ÷ 6 6 x=5 The value of x is 5.