Question

complete parts a–c for each quadratic equation.
a. find the value of the discriminant.
b. describe the number and type of roots.
c. find the exact solutions by using the quadratic formula.
$3x^2 + 9x - 2 = 0$
\bigcirc $105; 2$ irrational; $\dfrac{-9 \pm \sqrt{105}}{6}$
\bigcirc $\sqrt{105}; 2$ irrational; $\dfrac{-9 \pm \sqrt{105}}{6}$
\bigcirc $105; 2$ rational; $\dfrac{-9 \pm \sqrt{105}}{6}$
\bigcirc $57$ irrational; $\dfrac{-9 \pm \sqrt{57}}{6}$

Explanation:

Step 1: Recall the discriminant formula

For a quadratic equation \(ax^{2}+bx + c = 0\), the discriminant \(D=b^{2}-4ac\). In the equation \(3x^{2}+9x - 2=0\), we have \(a = 3\), \(b = 9\), and \(c=- 2\).
So, \(D=(9)^{2}-4\times3\times(-2)=81 + 24=105\).

Step 2: Analyze the nature of roots

Since the discriminant \(D = 105>0\) and \(\sqrt{105}\) is an irrational number (because \(105\) is not a perfect square), the quadratic equation will have two distinct irrational roots.

Step 3: Apply the quadratic formula

The quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Substituting \(a = 3\), \(b = 9\), and \(D = 105\) (so \(\sqrt{b^{2}-4ac}=\sqrt{105}\)) into the formula, we get:
\(x=\frac{-9\pm\sqrt{105}}{2\times3}=\frac{-9\pm\sqrt{105}}{6}\)

Answer:

\(105\); \(2\) irrational; \(\boldsymbol{\frac{-9\pm\sqrt{105}}{6}}\) (corresponding to the first option: \(105; 2\) irrational; \(\frac{-9\pm\sqrt{105}}{6}\))