Q. \(x^{2}+2x-8=0\)

Q: What are the solutions to x^2+2x-8=0?

\bigl(x=2\bigr) or \bigl(x=-4\bigr).

Q: How do you factor x^2+2x-8 to solve it?

Look for numbers with product -8 and sum 2: (x+4)(x-2)=0. Hence x=-4 or x=2.

Q: What does the quadratic formula give?

For a=1,b=2,c=-8: x=\dfrac{-2\pm\sqrt{2^2-4(1)(-8)}}{2}=\dfrac{-2\pm\sqrt{36}}{2}=\dfrac{-2\pm6}{2}, so x=2,-4.

Q: What is the discriminant, and what does it tell you?

\Delta=b^2-4ac=2^2-4(1)(-8)=4+32=36. Since \Delta>0, there are two distinct real solutions.

Q: How can you check the solutions quickly?

Substitute x=2: 4+4-8=0. Substitute x=-4: 16-8-8=0. Both satisfy the equation.

Answer

We solve the quadratic \(x^2+2x-8=0\) by factoring.

Find two numbers whose product is \(-8\) and sum is \(2\). Those numbers are \(4\) and \(-2\).

\[
x^2+2x-8=(x+4)(x-2)=0
\]

So \(x+4=0\) or \(x-2=0\).

\(x=-4\) or \(x=2\).

Detailed Explanation

We want to solve the equation

\[
x^2+2x-8=0
\]

Step 1: Identify the coefficients.

The equation is in the form \(ax^2+bx+c=0\), where:

\[
a=1,\quad b=2,\quad c=-8
\]

Step 2: Use the quadratic formula.

The quadratic formula is:

\[
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
\]

Step 3: Substitute \(a\), \(b\), and \(c\) into the formula.

\[
x=\frac{-2\pm\sqrt{(2)^2-4(1)(-8)}}{2(1)}
\]

Step 4: Simplify inside the square root (the discriminant).

\[
(2)^2=4
\]

\[
-4(1)(-8)=+32
\]

So the discriminant becomes:

\[
b^2-4ac=4+32=36
\]

Step 5: Take the square root of the discriminant.

\[
\sqrt{36}=6
\]

Step 6: Substitute back and simplify.

\[
x=\frac{-2\pm 6}{2}
\]

Step 7: Compute the two solutions.

First solution using \(+\):

\[
x=\frac{-2+6}{2}=\frac{4}{2}=2
\]

Second solution using \(-\):

\[
x=\frac{-2-6}{2}=\frac{-8}{2}=-4
\]

Final Answer: The solutions are

\[
x=2 \quad \text{or} \quad x=-4
\]

See full solution

Graph

Need help with homework? Try our AI tools now!

Homework AI

Algebra FAQ

What are the solutions to \(x^2+2x-8=0\)?

\(\bigl(x=2\bigr)\) or \(\bigl(x=-4\bigr)\).

How do you factor \(x^2+2x-8\) to solve it?

Look for numbers with product \(-8\) and sum \(2\): \((x+4)(x-2)=0\). Hence \(x=-4\) or \(x=2\).

What does the quadratic formula give?

For \(a=1,b=2,c=-8\): \(x=\dfrac{-2\pm\sqrt{2^2-4(1)(-8)}}{2}=\dfrac{-2\pm\sqrt{36}}{2}=\dfrac{-2\pm6}{2}\), so \(x=2,-4\).

Is there a way to solve by completing the square?

\(x^2+2x-8=0\Rightarrow x^2+2x=8\Rightarrow (x+1)^2-1=8\Rightarrow (x+1)^2=9\Rightarrow x=-1\pm3\), so \(x=2,-4\).

What is the discriminant, and what does it tell you?

\(\Delta=b^2-4ac=2^2-4(1)(-8)=4+32=36\). Since \(\Delta>0\), there are two distinct real solutions.

How can you check the solutions quickly?

Substitute \(x=2\): \(4+4-8=0\). Substitute \(x=-4\): \(16-8-8=0\). Both satisfy the equation.

Try solving x^2+2x-8=0.
Add step-by-step help with tools.

298,376+ active customers
Math, Geometry, Trigonometry, etc.

Math Solver Geometry Solver Trigonometry Solver