Question

which is a solution to the equation?
$(x - 3)(x - 5) = 35$
$\bigcirc$ $x = -8$
$\bigcirc$ $x = -5$
$\bigcirc$ $x = 2$
$\bigcirc$ $x = 10$

Explanation:

Step1: Expand the left side

First, expand \((x - 3)(x - 5)\) using the FOIL method.
\((x - 3)(x - 5)=x^2-5x-3x + 15=x^2-8x + 15\)
So the equation becomes \(x^2-8x + 15=35\)

Step2: Rearrange to standard quadratic form

Subtract 35 from both sides: \(x^2-8x+15 - 35=0\), which simplifies to \(x^2-8x - 20=0\)

Step3: Test each option

• For \(x=-8\): Substitute into \(x^2-8x - 20\), we get \((-8)^2-8\times(-8)-20=64 + 64-20 = 108

eq0\)

• For \(x=-5\): Substitute into \(x^2-8x - 20\), we get \((-5)^2-8\times(-5)-20=25 + 40-20 = 45

eq0\)

• For \(x = 2\): Substitute into \(x^2-8x - 20\), we get \(2^2-8\times2-20=4-16 - 20=-32

eq0\)

• For \(x = 10\): Substitute into \(x^2-8x - 20\), we get \(10^2-8\times10-20=100-80 - 20 = 0\)

Answer:

D. \(x = 10\)