Question

which equation is true for x = -6 and x = 2?
○ 2x² - 16x + 12 = 0
○ 2x² + 8x - 24 = 0
○ 3x² - 4x - 12 = 0
○ 3x² + 12x + 36 = 0

Explanation:

Step1: Recall Factor Theorem

If \( x = a \) and \( x = b \) are roots of a quadratic equation \( ax^2+bx + c=0 \), then the equation can be written as \( k(x - a)(x - b)=0 \), where \( k
eq0 \). Here, \( a=-6 \) and \( b = 2 \), so the equation is \( k(x + 6)(x - 2)=0 \).

Step2: Expand the equation

First, expand \( (x + 6)(x - 2) \):

$$
LATEXBLOCK0
$$

Multiply by \( k = 2 \) (to match the coefficient of \( x^2 \) in some options), we get \( 2(x^2 + 4x - 12)=2x^2+8x - 24 = 0 \).
We can also verify by substituting \( x=-6 \) and \( x = 2 \) into each option:

• For option \( 2x^2-16x + 12 = 0 \):
• When \( x=-6 \): \( 2(-6)^2-16(-6)+12=2\times36 + 96+12=72 + 96+12=180

eq0 \)

• For option \( 2x^2+8x - 24 = 0 \):
• When \( x=-6 \): \( 2(-6)^2+8(-6)-24=2\times36-48 - 24=72-48 - 24=0 \)
• When \( x = 2 \): \( 2(2)^2+8(2)-24=8 + 16-24=0 \)
• For option \( 3x^2-4x - 12 = 0 \):
• When \( x=-6 \): \( 3(-6)^2-4(-6)-12=3\times36 + 24-12=108 + 24-12=120

eq0 \)

• For option \( 3x^2+12x + 36 = 0 \):
• When \( x=-6 \): \( 3(-6)^2+12(-6)+36=3\times36-72 + 36=108-72 + 36=72

eq0 \)

Answer:

\( 2x^2 + 8x - 24 = 0 \) (the second option)