Q. Solve the quadratic equation for all values of x in simplest form: \[5\left(x^{2}-8\right)-9=6.\]

Q: What is the first simplification step for 5(x^2-8)-9=6?

Expand and combine constants: 5x^2-40-9=6, so 5x^2-49=6. Then isolate: 5x^2=55 and x^2=11..

Q: How do I get the values of x from x^2=11?.

Take square roots: x=\pm\sqrt{11}.

Q: Can I factor the equation instead of isolating x^2?

You can rewrite x^2-11=0 as (x-\sqrt{11})(x+\sqrt{11})=0, but isolating x^2 was simpler here.

Q: Is \sqrt{11} in simplest radical form?

Yes. 11 is prime, so \sqrt{11} cannot be simplified further.

Q: Could there be extraneous roots introduced by taking square roots?

No extraneous roots arise when solving x^2=a by taking \pm\sqrt{a}. If a>0 you get two real roots; a=0 gives one; a<0 gives no real roots.

Q: If the leading coefficient weren’t 5, would the method change?

No. Combine like terms, divide by the coefficient of x^2 to get x^2= constant, then take \pm square roots.

Answer

\[
5(x^2-8)-9=6
\]
\[
5x^2-40-9=6
\]
\[
5x^2-55=0
\]
\[
x^2=11
\]
\[
x=\pm\sqrt{11}
\]

Detailed Explanation

Write the given equation exactly:\[5(x^{2}-8)-9=6\]
Distribute 5 across the parentheses (multiply each term inside by 5):\[5\cdot x^{2}+5\cdot(-8)-9=6\]
So we get

\[5x^{2}-40-9=6\]
Combine the constant terms on the left side, \(-40-9=-49\):\[5x^{2}-49=6\]
Isolate the quadratic term by adding 49 to both sides:\[5x^{2}=6+49\]
Evaluate the sum on the right:

\[5x^{2}=55\]
Divide both sides by 5 to solve for \(x^{2}\):\[x^{2}=\dfrac{55}{5}=11\]
Take the square root of both sides. Remember both the positive and negative roots must be taken for a quadratic:\[x=\pm\sqrt{11}\]
State the solutions in simplest form:\[x=\sqrt{11}\quad\text{or}\quad x=-\sqrt{11}\]

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Algebra FAQs

What is the first simplification step for \(5(x^2-8)-9=6\)?

Expand and combine constants: \(5x^2-40-9=6\), so \(5x^2-49=6\). Then isolate: \(5x^2=55\) and \(x^2=11\)..

How do I get the values of \(x\) from \(x^2=11\)?.

Take square roots: \(x=\pm\sqrt{11}\).

Are both \(+\sqrt{11}\) and \(-\sqrt{11}\) valid solutions?.

Yes. Substituting either into the original equation satisfies it, because squaring removes the sign..

Can I factor the equation instead of isolating \(x^2\)?

You can rewrite \(x^2-11=0\) as \((x-\sqrt{11})(x+\sqrt{11})=0\), but isolating \(x^2\) was simpler here.

Is \( \sqrt{11} \) in simplest radical form?

Yes. 11 is prime, so \(\sqrt{11}\) cannot be simplified further.

What is the decimal approximation of the solutions?

Could there be extraneous roots introduced by taking square roots?

No extraneous roots arise when solving \(x^2=a\) by taking \(\pm\sqrt{a}\). If \(a>0\) you get two real roots; \(a=0\) gives one; \(a<0\) gives no real roots.

If the leading coefficient weren’t 5, would the method change?

No. Combine like terms, divide by the coefficient of \(x^2\) to get \(x^2=\) constant, then take \(\pm\) square roots.

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