Question

Solve the radical equation to find the two solutions. Round to one decimal place if needed.
^3 square root of 2x^2+19x+17=2

Answer 1

To solve the radical equation:

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

we begin by eliminating the cube root on the left side. We do this by cubing both sides of the equation:

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

Calculating \(2^3\):

\[
2^3 = 8
\]

So the equation simplifies to:

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

Next, we subtract 8 from both sides to set the equation to zero:

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

This simplifies to:

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

Now, we can use the quadratic formula to solve for \(x\), where \(a = 2\), \(b = 19\), and \(c = 9\):

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

First, we calculate the discriminant:

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

Calculating \(19^2\):

\[
19^2 = 361
\]

Calculating \(4 \cdot 2 \cdot 9\):

\[
4 \cdot 2 \cdot 9 = 72
\]

Now subtract:

\[
361 - 72 = 289
\]

The discriminant is \(289\). Now we can find the solutions using the quadratic formula:

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

Since \( \sqrt{289} = 17 \), we have:

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

Calculating both potential solutions:

1. First solution:

\[
x = \frac{-19 + 17}{4} = \frac{-2}{4} = -\frac{1}{2} = -0.5
\]

2. Second solution:

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

The two solutions are:

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

Thus, the two solutions are:

\[
\boxed{-0.5} \quad \text{and} \quad \boxed{-9}
\]