Question

what are the solutions of the equation $4x^2 + 3x = 24 - x$?

• $-3$, $2$, or $4$
• $-3$ or $2$
• $-2$, $3$, or $4$
• $-2$ or $3$

Explanation:

Step1: Rearrange the equation

First, we need to rearrange the given equation \(4x^{2}+3x = 24 - x\) into standard quadratic form \(ax^{2}+bx + c=0\).
Add \(x\) to both sides and subtract 24 from both sides:
\(4x^{2}+3x+x - 24=0\)
Simplify the like terms:
\(4x^{2}+4x - 24 = 0\)
We can divide the entire equation by 4 to simplify:
\(x^{2}+x - 6=0\)

Step2: Factor the quadratic equation

We need to factor the quadratic equation \(x^{2}+x - 6 = 0\). We look for two numbers that multiply to \(- 6\) and add up to \(1\). The numbers are \(3\) and \(-2\) since \(3\times(-2)=-6\) and \(3+(-2) = 1\).
So, the factored form is \((x + 3)(x - 2)=0\)

Step3: Solve for x

Using the zero - product property, if \(ab = 0\), then either \(a = 0\) or \(b = 0\).
For \((x + 3)(x - 2)=0\), we have:
Case 1: \(x+3=0\), then \(x=-3\)
Case 2: \(x - 2=0\), then \(x = 2\)

Answer:

\(-3\) or \(2\) (corresponding to the option: \(-3\) or \(2\))