Q. \(x^2+6x+8=0\)

Q: What is the discriminant b^2-4ac for this equation?

For a=1,b=6,c=8: \Delta=6^2-4\cdot1\cdot8=36-32=4. Since \Delta>0, there are two real roots.

Q: How do you find two numbers that multiply to 8 and add to 6?

(2,4) work because 2\cdot4=8 and 2+4=6. So x^2+6x+8=(x+2)(x+4).

Q: Check the solutions in the original equation.

For x=-2: 4-12+8=0. For x=-4: 16-24+8=0. Both satisfy the equation.

Answer

To solve \(x^2+6x+8=0\),factor the quadratic.

\(x^2+6x+8=(x+2)(x+4)=0\).

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

Final result: \(x=-2,\,-4\).

Detailed Explanation

We want to solve the quadratic equation

\[
x^{2} + 6x + 8 = 0
\]

Step 1: Factor the quadratic.

To factor \(x^{2} + 6x + 8\), we look for two numbers that:

Multiply to \(8\)
Add to \(6\)

The numbers \(2\) and \(4\) work because

\[
2 \cdot 4 = 8 \quad \text{and} \quad 2 + 4 = 6
\]

So the quadratic factors as

\[
x^{2} + 6x + 8 = (x + 2)(x + 4)
\]

Step 2: Set each factor equal to zero.

The equation becomes

\[
(x + 2)(x + 4) = 0
\]

By the zero product property, we must have either

\[
x + 2 = 0
\]
or
\[
x + 4 = 0
\]

Step 3: Solve each linear equation.

First:

\[
x + 2 = 0
\]

Subtract \(2\) from both sides:

\[
x = -2
\]

Second:

\[
x + 4 = 0
\]

Subtract \(4\) from both sides:

\[
x = -4
\]

Final Answer:

The solutions are

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

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Algebra FAQ

Solve \(x^2+6x+8=0\) using factoring.

\(x^2+6x+8=(x+2)(x+4)=0\Rightarrow x=-2,-4.\)

What are the roots using the quadratic formula?

\(x=\frac{-6\pm\sqrt{36-32}}{2}=\frac{-6\pm 2}{2}\Rightarrow x=-2,-4.\)

Can you complete the square for \(x^2+6x+8=0\)?

\(x^2+6x+8=0\Rightarrow (x+3)^2-1=0\Rightarrow (x+3)^2=1\Rightarrow x=-4,-2.\)

What is the discriminant \(b^2-4ac\) for this equation?

For \(a=1,b=6,c=8\): \(\Delta=6^2-4\cdot1\cdot8=36-32=4.\) Since \(\Delta>0\), there are two real roots.

How do you find two numbers that multiply to \(8\) and add to \(6\)?

\((2,4)\) work because \(2\cdot4=8\) and \(2+4=6.\) So \(x^2+6x+8=(x+2)(x+4).\)

Check the solutions in the original equation.

For \(x=-2\): \(4-12+8=0.\) For \(x=-4\): \(16-24+8=0.\) Both satisfy the equation.

Test solve: x²+6x+8=0.
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