If k(x)= 2x^2+6x-4 and w(x)=-3x^2-18x+32, find all real x-values such that k(x)= w(x)

Answer 1

Since you're looking for all values where k(x) = w(x), set the two equations equal to each other and solve for x.

Bring everything to 1 side since we have a quadratic

2x^2 + 6x - 4 = -3x^2 - 18x + 32

5x^2 -24x - 36 = 0

Use quadratic formula to solve for x

-(-24)±√((-24)² - 4(5)(-36)) ⁄ 2(5) = x

(24±√(1296)) ⁄ 10 = x

(24 + 36) / 10 = x and (24 - 36) / 10 = x

-6 = x and 1.2 = x

Plug these in to the original formulas to get respective y values

k(x) = 2x^2 + 6x - 4

k(x) = 2(-6)^2 + 6(-6) - 4

k(x) = 32

k(x) = 2x^2 + 6x - 4

k(x) = 2(1.2)^2 + 6(1.2) - 4

k(x) = 6.08

The real values for which k(x) = w(x) are:

(-6,32) and (1.2,6.08)

Answer 2

2x² + 6x - 4 = -3x² - 18x + 32

5x² + 24x - 36 = 0

(5x - 6)(x + 6) = 0

What are the two solutions?