Question 40
Solve the equation using the quadratic formula.
\[
3 x^{2}+x-5=0
\]
(A)
B
\[
\left\{\frac{-1-\sqrt{61}}{2}, \frac{-1+\sqrt{61}}{2}\right\}
\]
(C)
\[
\left\{\frac{1-\sqrt{61}}{6}, \frac{1+\sqrt{61}}{6}\right\}
\]
(D)
\[
\left\{\frac{-1-\sqrt{61}}{6}, \frac{-1+\sqrt{61}}{6}\right\}
\]

Answer

Expert–verified

Hide Steps

Answer

\(\boxed{\left\{\frac{-1-\sqrt{61}}{6}, \frac{-1+\sqrt{61}}{6}\right\}}\)

Steps

Step 1 ：Given the quadratic equation: \(3x^2 + x - 5 = 0\)

Step 2 ：Use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Step 3 ：Plug in the values: a = 3, b = 1, and c = -5

Step 4 ：Calculate the discriminant: \(b^2 - 4ac = 1^2 - 4(3)(-5) = 61\)

Step 5 ：Plug the discriminant into the quadratic formula: \(x = \frac{-1 \pm \sqrt{61}}{2(3)}\)

Step 6 ：Simplify the expression: \(x = \frac{-1 \pm \sqrt{61}}{6}\)

Step 7 ：\(\boxed{\left\{\frac{-1-\sqrt{61}}{6}, \frac{-1+\sqrt{61}}{6}\right\}}\)