Question

question 18 of 25
which of the following are solutions to the quadratic equation? check all that apply.
$2x^2 + 9x - 3 = -x^2 + x$

a. $\frac{1}{3}$
b. $-3$
c. $3$
d. $-\frac{1}{3}$
e. $-2$
f. $1$

Explanation:

Step1: Simplify the equation

First, we need to get all terms on one side of the equation. Add \(x^{2}\) to both sides and subtract \(x\) from both sides:
\[

$$\begin{align*} 2x^{2}+9x - 3+x^{2}-x&=-x^{2}+x+x^{2}-x\\ 3x^{2}+8x - 3&=0 \end{align*}$$

\]

Step2: 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, \(a = 3\), \(b = 8\), and \(c=-3\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(8)^{2}-4\times3\times(-3)=64 + 36 = 100\).
Then, find the roots:
\[

$$\begin{align*} x&=\frac{-8\pm\sqrt{100}}{2\times3}\\ &=\frac{-8\pm10}{6} \end{align*}$$

\]
For the plus sign: \(x=\frac{-8 + 10}{6}=\frac{2}{6}=\frac{1}{3}\)
For the minus sign: \(x=\frac{-8-10}{6}=\frac{-18}{6}=-3\)

Answer:

A. \(\frac{1}{3}\), B. \(-3\)