To determine which of the given options are solutions to the equation [tex]\( 2x^2 + 14x - 4 = -x^2 + 3x \)[/tex], follow these steps:
1. Simplify the equation:
Start by getting all terms on one side of the equation:
[tex]\[
2x^2 + 14x - 4 + x^2 - 3x = 0
\][/tex]
Combine like terms:
[tex]\[
(2x^2 + x^2) + (14x - 3x) - 4 = 0
\][/tex]
[tex]\[
3x^2 + 11x - 4 = 0
\][/tex]
2. Solve the quadratic equation:
The standard quadratic formula:
[tex]\[
ax^2 + bx + c = 0
\][/tex]
Where [tex]\( a = 3 \)[/tex], [tex]\( b = 11 \)[/tex], and [tex]\( c = -4 \)[/tex]. Use the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Substitute [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[
x = \frac{-11 \pm \sqrt{11^2 - 4 \cdot 3 \cdot (-4)}}{2 \cdot 3}
\][/tex]
Simplify under the square root:
[tex]\[
x = \frac{-11 \pm \sqrt{121 + 48}}{6}
\][/tex]
[tex]\[
x = \frac{-11 \pm \sqrt{169}}{6}
\][/tex]
[tex]\[
x = \frac{-11 \pm 13}{6}
\][/tex]
So, we get two solutions for [tex]\( x \)[/tex]:
[tex]\[
x = \frac{-11 + 13}{6} = \frac{2}{6} = \frac{1}{3}
\][/tex]
and
[tex]\[
x = \frac{-11 - 13}{6} = \frac{-24}{6} = -4
\][/tex]
So the solutions to the quadratic equation [tex]\( 3x^2 + 11x - 4 = 0 \)[/tex] are [tex]\( x = \frac{1}{3} \)[/tex] and [tex]\( x = -4 \)[/tex].
3. Check the given options:
Now, let's see which of the provided options match these solutions:
- A. 1 -> Not a solution
- B. 4 -> Not a solution
- C. [tex]\(-\frac{1}{2}\)[/tex] -> Not a solution
- D. [tex]\(\frac{1}{3}\)[/tex] -> Solution
- E. 3 -> Not a solution
- F. -4 -> Solution
Therefore, the correct answers are options D ([tex]\(\frac{1}{3}\)[/tex]) and F (-4).
