Question

which is a solution to the equation?
$(x - 2)(x + 5) = 18$
$\bigcirc$ $x = -10$
$\bigcirc$ $x = -7$
$\bigcirc$ $x = -4$
$\bigcirc$ $x = -2$

Explanation:

Step1: Expand the left side

First, we expand \((x - 2)(x + 5)\) using the distributive property (FOIL method).
\((x - 2)(x + 5)=x^2 + 5x - 2x - 10=x^2 + 3x - 10\)
So the equation becomes \(x^2 + 3x - 10 = 18\).

Step2: Rearrange to standard quadratic form

Subtract 18 from both sides to set the equation to zero:
\(x^2 + 3x - 10 - 18 = 0\)
\(x^2 + 3x - 28 = 0\)

Step3: Factor the quadratic (or use quadratic formula)

We can factor \(x^2 + 3x - 28\) as \((x + 7)(x - 4)=0\) (since \(7\times(- 4)=-28\) and \(7-4 = 3\)).
Setting each factor equal to zero gives \(x+7 = 0\) or \(x - 4=0\), so \(x=-7\) or \(x = 4\).

Alternatively, we can test each option by substituting into the original equation:

• For \(x=-10\): \((-10 - 2)(-10 + 5)=(-12)\times(-5)=60

eq18\)

• For \(x=-7\): \((-7 - 2)(-7 + 5)=(-9)\times(-2)=18\), which works.
• For \(x=-4\): \((-4 - 2)(-4 + 5)=(-6)\times(1)=-6

eq18\)

• For \(x=-2\): \((-2 - 2)(-2 + 5)=(-4)\times(3)=-12

eq18\)

Answer:

\(x=-7\) (the option is \(x = -7\))