Page 146 - Algebra

4.
4+5 𝑥2−4 (𝑥+2)2
Simplify the polynomials
=4+5 𝑥2−4 (𝑥+2)2
=4+ (𝑥−2)(𝑥+2)
5
=1( 𝑥+2
= 𝑥+2( =1(
𝑥+2
= 1 (9𝑥−2)
(𝑥+2)(𝑥+2)
4+5) (𝑥−2) (𝑥+2)
The LCD of the equation is (x – 2) (x + 2)
1
4𝑥+8+5𝑥−10 ) (𝑥−2)(𝑥+2)
9𝑥−2 ) (𝑥−2)(𝑥+2)
𝑥+2 𝑥2−4 = 9𝑥−2
(𝑥+2)(𝑥2−4)
5. Cross multiply the equations to solve (x – 3) (x – 4) = 3x (x + 1)
x2 – 3x – 4x + 12= 3x2 + 3x
–2x2 – 7x + 12 = 3x
–2x2 – 10x + 12 = 0 2x2 +10x–12=0
x2 +5x–6=0
x2 + 6x – x – 6 = 0
x(x + 6) – 1(x + 6) = 0 (x – 1) (x + 6) = 0
x = 1 or –6
6. A fraction is undefined when the denominator is 0. So, equate the denominator to 0
(𝑥+𝑘)2 −2(𝑥+𝑘)−3=0 (3+𝑘)2 −2(3+𝑘)−3=0 9 + 6k + k2 – 6 – 2k − 3 = 0 4k + k2 = 0
k(k + 4) = 0 k = 0 or – 4
Page 145 of 177
Algebra I & II