Day 6 Solving by Factoring

University:
California State University, Northridge
Course:
ACCT 380 | Cost Accounting
Academic year:

2024
Views:

0

Pages:

1
Author:

gaurav boparai

Day 6: Solving Quadratics by Factoring Name: ____________________________ 3 Period: __________ Factor the quadratic, then calculate the x-intercepts. 1. 2𝑥 2 + 𝑥 − 6 = 0 2x^2+4x3x6=0 (2x^2+4x) + (3x6) =0 2x(x+2) 3(x+2) =0 (2x3) (x+2) =0 2x3=0 x=3/2 x+2=0 x=-2 2. 4𝑥 2 − 5𝑥 + 1 = 0 6𝑥 2 + 4𝑥 + 6 = −3𝑥 + 4 4x^24x1x+1=0 (4x^24x) + (1x+1) =0 4x(x1) 1(x1) =0 (4x1) (x1) =0 4x1=0 x= 4/1 Step #1: Rewrite the equation in Standard Form by getting the equation to equal zero. x1=0 x=1 6𝑥 2 + 7𝑥 + 2 = 0 Step #2: Factor the quadratic. Solve for 𝑥 by factoring. (2𝑥 + 1)(3𝑥 + 2) = 0 3. 7𝑥 2 + 5𝑥 = 2 Step #3: Set each factor equal to zero, then solve for x. 7x^2+5x2=0 7x^2+7x2x2=0 (7x^2+7x) + (2x2) =0 7x(x+1) 2(x+1) =0 (7x2) (x+1) =0 2𝑥 + 1 = 0 3𝑥 + 2 = 0 1 7x2=0 x=7/2 4. 6𝑥 2 + 2𝑥 = 3𝑥 + 2 5. 2𝑥 2 − 10𝑥 − 3 = 𝑥 − 8 6x^2+2x3x2=0 6^2 2 =0 6x^2 +3x 4x 2 =0 (6x^2 +3x) + (4x 2) =0 3x(2x+1) 2(2x+1) =0 (2x+1) (3x2) =0 2x+1=0 x= 2/1 2 𝑥 = −2 x+1=0 x=1 3x2=0 x= 3/2 6. −12𝑥 2 − 1 = −2𝑥 2 − 19𝑥 + 5 2x^2 10x3x+8 =0 2x^2 11x+5 =0 2x^2 10xx+5 =0 (2x^2 10x) + (x+5) =0 2x(x5) 1(x5) =0 (2x1) (x5) =0 2x1=0 x=1/2 𝑥 = −3 12x^2 +2x^2 +19x15=0 10x^2 +19x6=0 10x^2 15x4x+6=0 (10x^2 15x) + (4x+6) =0 5x(2x3) 2(2x3) =0 (5x2) (2x3) =0 x5=0 x=5 5x2=0 x= 2/5 2x3=0 x=3/2 Mixed Review: Solve for 𝑥. Show your work. 7. (𝑥 − 1)2 + 4 = 13 (x1)^2 =134 (x1)^2 =9 x1=±3 x1=3 =3+1 x=4 x1=3 x=3+1 x=2 8. 4𝑥 2 + 1 = 10 4x^2 =101 4x^2 =9 x^2 = 9/4 x=± 4/9 x= ±2/3 x= 2/3 x= -3/2 9. 1 3 (𝑥 + 2)2 − 1 = 2 1/3(x+2)^2 =2+1 1(x=2)^2 =3*3 (x=2)^2 =9 x+2=±3 x+2=±3 x=32 x=1 x+2=3 x=32 x= -5