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)
x=

Answer 1

To solve the equation

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

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

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

Calculating \(2^3\):

\[
2^3 = 8.
\]

Now we have the equation:

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

Next, we subtract 8 from both sides:

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

which simplifies to:

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

Now we can use the quadratic formula to find the solutions for \(x\). The quadratic formula is:

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

where \(a = 2\), \(b = 19\), and \(c = 9\).

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 we take the square root of the discriminant:

\[
\sqrt{289} = 17.
\]

Now we can plug this back into the quadratic formula:

\[
x = \frac{-19 \pm 17}{2 \cdot 2}.
\]

This gives us two potential solutions:

1. First solution:

\[
x = \frac{-19 + 17}{4} = \frac{-2}{4} = -0.5.
\]

2. Second solution:

\[
x = \frac{-19 - 17}{4} = \frac{-36}{4} = -9.
\]

Therefore, the two solutions are

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

Rounding to one decimal place if needed:

\[
x = -0.5, \quad -9.
\]

So the final answers are:

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