Question

What is the solution set of x2 – 10 = 30x?

{–220, 250}
{–250, 220}
{15 – StartRoot 235 EndRoot, 15 + StartRoot 235 EndRoot}
{–15 – StartRoot 235 EndRoot, –15 + StartRoot 235 EndRoot}

Answer 1

To solve the equation \( x^2 - 10 = 30x \), we can first rearrange it into standard quadratic form:

\[
x^2 - 30x - 10 = 0
\]

Now, we can apply the quadratic formula, which is given by:

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

In our equation, \( a = 1 \), \( b = -30 \), and \( c = -10 \). Plugging these values into the formula gives us:

\[
x = \frac{-(-30) \pm \sqrt{(-30)^2 - 4 \cdot 1 \cdot (-10)}}{2 \cdot 1}
\]

Calculating \( b^2 - 4ac \):

\[
(-30)^2 = 900
\]
\[
4 \cdot 1 \cdot (-10) = -40
\]
\[
b^2 - 4ac = 900 + 40 = 940
\]

Now, substituting this back into the quadratic formula:

\[
x = \frac{30 \pm \sqrt{940}}{2}
\]

We simplify \( \sqrt{940} \):

\[
\sqrt{940} = \sqrt{4 \cdot 235} = 2\sqrt{235}
\]

Now substitute back:

\[
x = \frac{30 \pm 2\sqrt{235}}{2}
\]

This simplifies to:

\[
x = 15 \pm \sqrt{235}
\]

Thus, the solutions are:

\[
x = 15 - \sqrt{235} \quad \text{and} \quad x = 15 + \sqrt{235}
\]

From your options, the correct solution set is:

\[
\{ 15 - \sqrt{235}, 15 + \sqrt{235} \}
\]