University:
College Entrance Preparation
Course:
ACT Preparation (Mathematics)
Academic year:

2021
Views:

13

Pages:

3
Author:

Tyshawn Shaffer

1.5 Chpt REVIEW - ANSWERS Factor completely 1. x2 – 5x – 24 Must: Multiply to -24 Add to -5 2. 5x2 –9x – 2 5x2 -10x +1x -2 (5x2 -10x) (+1x -2) 5x(x -2) +1(1x -2) (5x+1)(x-2) (x-8)(x+3) -10x2 -10x +1x -9x Complete the square and then factor 3. x2 – 16x + 64= (x + 8)2 4. x2 + x + (1/4) = (x + ½)2 To complete the square, add (b/2)2 Solve using factoring 5. x2 + 6x – 16 = 0 Must: Multiply to -16 Add to -6 (x+2)(x-8)=0 Z.P.P. x+2=0 OR x-8=0 x = -2 OR x = -8 6. 7x2 - x – 18 = (x+1)(x-3) 7x2 - x – 18 = x2 -2x -3 6x2 +x -15 = 0 6x2 +10x -9x -15 = 0 (6x2 +10x) (-9x -15 )= 0 2x (3x +5) -3 (3x +5 )= 0 (2x-3)(3x+5) = 0 ZPP 2x-3 = 0 OR 3x+5=0 x= 3/2 OR x= -5/3 Solve using the square root property 7. 3x2 = - 36 X2 = -12 X= √-12 X= ±2√3 i 8. (x – 8)2 = 25 x-8 = ±5 x = 8+5 Or x= 8-5 x=13 or x= 3 Solve by completing the square 9. x2 + 7x + 12 = 0 x2 +7x = -12 x2 +7x + (7/2)2 = -12 + (7/2)2 (x + 7/2 )2 = ¼ x + 7/2 = ± ½ x = -7/2 ± ½ x = -7/2 +1/2 = -6/2 = -3 OR x= -7/2 – 1/2 = -8/2 = -4 x = -3 OR x=-4 10. 2x2 - 12x - 32 = 0 x2 -6x -16 = 0 x2 -6x =16 2 x -6x +9 =16 +9 (x – 3)2 = 25 x–3 =±5 x= 3 ± 5 x = 3+5 or x= 3-5 x = 8 OR x = -2 11. 9x2 + 79 = -18x 9x2 +18x +79 = 0 12. 7x2 -12x = 2x + 56 7x2 – 14x -56 = 0 -90x2 10x -9x 1x 9x2 + 18x = -79 2 9(x + 2x ) = -79 9(x2+ 2x +1 ) = -79 + 9(1) 9(x+1)2 = -70 (x+1)2 = -70/9 x2 -2x -8 = 0 x2 -2x =8 x2 -2x +1 = 8+1 (x-1)2 = 9 x-1 = ±3 √70 𝑖 3 √70 -1 ± 3 𝑖 x+1 = ± x = Solve by using the quadratic formula 13. x2 + 4x – 21 = 0 𝑥= 𝑥= 𝑥= 𝑥= −𝑏±√𝑏 2 −4𝑎𝑐 2𝑎 −4±√42 −4(1)(−21) 2(1) −4±√100 −4±10 𝑥= 𝑥= −4±√42 −4(2)(1) 2(2) −4±√8 4 −4±2√2 𝒙= −𝑏±√𝑏 2 −4𝑎𝑐 2𝑎 1±√12 −4(4)(−3) 2(4) 1±√49 8 1±7 8 x= 1 or -3/4 Solve using any method 15. 4x2 = -8x-2 4x2 +8x +2 = 0 2x2 +4x +1 = 0 can’t factor. Use quad form. 𝑥= 14. 4x2 = x + 3 4x2 – x -3 = 0 𝑥= 2 𝑥= x = 4 or x = -2 𝑥= 2 x= 3 or x=-7 𝑥= x= 1± 3 4 −𝟐±√𝟐 𝟐 16. 2 (3x + 1)2 -8 = 280 2 (3x + 1)2 = 288 (3x + 1)2 = 144 3x + 1 = ±12 x = (1±12) / 3 x= 13/3 or x=-11/3 Compute the discriminant. Then determine the number and type of solutions: 1 real solution 2 real solutions 2 complex solutions 15. x2 = 12x + 4 x2-12x-4 = 0 a=1 b= -12 c=-4 (-12)2 – 4(1)(-4) = 144+16=160 Positive → 2 real solutions 16. 4x2 + 6x + 5 = 0 a=4 b= 6 c= 5 62 – 4(4)(5) = 36-80 = -44 Negative → 2 complex solutions 17. x2 – 16x + 64 = 0 a=1 b= -16 c= 64 (-16)2 – 4(1)(64) = 256-256 = 0 0 → 1 real solution