Q. For what values of \(x\) is \(x^2 + 2x = 24\) true?
Answer
Subtract 24:
\[
x^{2}+2x-24=0
\]
Factor:
\[
(x+6)(x-4)=0
\]
Hence
\[
\boxed{\,x=-6\quad\text{or}\quad x=4\,}
\]
Detailed Explanation
Problem: Solve: \( x^{2} + 2x = 24 \)
Step 1 – Rearrange to standard form
Subtract 24 from both sides:
\[
x^{2} + 2x – 24 = 0
\]
Step 2 – Factor the quadratic
Find two numbers whose product is \(-24\) and whose sum is \(2\): these are \(6\) and \(-4\). Factor the quadratic:
\[
x^{2} + 2x – 24 = (x + 6)(x – 4)
\]
Step 3 – Apply the zero product property and solve
Set each factor equal to zero:
\[
x + 6 = 0 \quad\text{or}\quad x – 4 = 0
\]
So
\[
x = -6 \quad\text{or}\quad x = 4
\]
Answer
\[
\{ -6,\, 4 \}
\]
See full solution
FAQs
What values of x satisfy (x^2 + 2x = 24)?
Move 24: (x^2+2x-24=0). Factor: ((x+6)(x-4)=0). So (x=-6) or (x=4).
Can I use the quadratic formula on (x^2+2x-24=0)?
Yes: \(x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-24)}}{2} = \frac{-2 \pm \sqrt{100}}{2} = \frac{-2 \pm 10}{2}\), giving 4 and -6.
How do I complete the square for (x^2+2x=24)?
Write ((x+1)^2-1=24), so ((x+1)^2=25). Hence (x+1=pm5), giving (x=4) or (x=-6).
How many real solutions does (x^2+2x=24) have?
Two distinct real solutions, because the discriminant \(b^2 - 4ac = 100 > 0\).
What is the graph of (y=x^2+2x-24)?
parabola opening upward (a=1), crossing the x-axis at (-6) and (4), with vertex at (x=-1, y=-25).
Are the solutions integers or rational numbers?
Are the solutions integers or rational numbers?
What are the sum and product of the roots?
For \(x^2+2x-24\), sum = \(-\frac{b}{a} = -2\) and product = \(\frac{c}{a} = -24\). Indeed \(-6 + 4 = -2\) and \((-6)(4) = -24\).
How can I check my solutions?
Substitute back: (4^2+2(4)=16+8=24); ((-6)^2+2(-6)=36-12=24). Both satisfy the original equation.
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