Question

Solve the radical equation 2x2+19x+17−−−−−−−−−−−−√3=2
to find the two solutions. Round the answer to one decimal place if needed.(1 point)

Answer 1

To solve the equation:

\[
\sqrt[3]{2x^2 + 19x + 17} = 2
\]

we start by eliminating the cube root by cubing both sides:

\[
2x^2 + 19x + 17 = 2^3
\]

Calculating \(2^3\):

\[
2x^2 + 19x + 17 = 8
\]

Now, we can rearrange this equation:

\[
2x^2 + 19x + 17 - 8 = 0
\]

which simplifies to:

\[
2x^2 + 19x + 9 = 0
\]

Now, we will use the quadratic formula to find the solutions, where \(a = 2\), \(b = 19\), and \(c = 9\):

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

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

\[
b^2 = 19^2 = 361
\]
\[
4ac = 4 \cdot 2 \cdot 9 = 72
\]
\[
b^2 - 4ac = 361 - 72 = 289
\]

Now substituting these values into the quadratic formula:

\[
x = \frac{-19 \pm \sqrt{289}}{2 \cdot 2}
\]

Calculating \(\sqrt{289}\):

\[
\sqrt{289} = 17
\]

Now substituting back into the formula:

\[
x = \frac{-19 \pm 17}{4}
\]

This gives us two cases:

1. \(x = \frac{-19 + 17}{4} = \frac{-2}{4} = -0.5\)
2. \(x = \frac{-19 - 17}{4} = \frac{-36}{4} = -9\)

Thus, the two solutions to the equation are:

\[
x = -0.5 \quad \text{and} \quad x = -9
\]

Rounding to one decimal place, we have:

\[
x \approx -0.5 \quad \text{and} \quad x = -9.0
\]

Final solutions:

\[
\boxed{-0.5} \quad \text{and} \quad \boxed{-9.0}
\]