Question

solve the radical equation \\(sqrt3{2x^2 + 19x + 17} = 2\\) to find the two solutions. round the answer to one decimal place if needed. (2 points)

\\(x = square, x = square\\)

Explanation:

Step1: Eliminate the cube root

To solve the equation \(\sqrt[3]{2x^{2}+19x + 17}=2\), we cube both sides of the equation. Cubing a cube root will cancel out the cube root operation. So we have:
\((\sqrt[3]{2x^{2}+19x + 17})^{3}=2^{3}\)
Which simplifies to:
\(2x^{2}+19x + 17 = 8\)

Step2: Rearrange into standard quadratic form

Subtract 8 from both sides of the equation to get a quadratic equation in the form \(ax^{2}+bx + c=0\).
\(2x^{2}+19x+17 - 8=0\)
\(2x^{2}+19x + 9=0\)

Step3: Solve the quadratic equation

We can use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for a quadratic equation \(ax^{2}+bx + c = 0\). Here, \(a = 2\), \(b=19\) and \(c = 9\).
First, calculate the discriminant \(\Delta=b^{2}-4ac\):
\(\Delta=(19)^{2}-4\times2\times9\)
\(=361 - 72\)
\(=289\)
Then, find the roots:
\(x=\frac{-19\pm\sqrt{289}}{2\times2}=\frac{-19\pm17}{4}\)

For the plus sign:
\(x_{1}=\frac{-19 + 17}{4}=\frac{-2}{4}=- 0.5\)

For the minus sign:
\(x_{2}=\frac{-19-17}{4}=\frac{-36}{4}=-9\)

Answer:

\(x=-9\), \(x = - 0.5\)