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$\frac{1}{2}c^{2}+3c=1$ $\newline$ Let $c=f$ and $c=g$ be the solutions to the given equation. If f > g, which of the following is the value of $f$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $-3+\sqrt{2}$ $\newline$ (B) $-3+\sqrt{11}$ $\newline$ (C) $3+\sqrt{2}$ $\newline$ (D) $3+\sqrt{11}$

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Full solution

Q.

\frac{1}{2}c^{2}+3c=1

\newline

Let

c=f

and

c=g

be the solutions to the given equation. If

f > g

, which of the following is the value of

f

?

\newline

Choose

1

answer:

\newline

(A)

-3+\sqrt{2}

\newline

(B)

-3+\sqrt{11}

\newline

(C)

3+\sqrt{2}

\newline

(D)

3+\sqrt{11}

Rewrite Equation: Rewrite the given equation in standard quadratic form. $\newline$ The given equation is $(\frac{1}{2})c^{2} + 3c = 1$ . To rewrite it in standard form, we need to have $0$ on one side of the equation. $\newline$ $(\frac{1}{2})c^{2} + 3c - 1 = 0$ $\newline$ Multiply through by $2$ to get rid of the fraction: $\newline$ $c^{2} + 6c - 2 = 0$
Use Quadratic Formula: Use the quadratic formula to find the solutions for $c$ . The quadratic formula is $c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = 1$ , $b = 6$ , and $c = -2$ in our equation. $c = \frac{-6 \pm \sqrt{6^2 - 4(1)(-2)}}{2(1)}$ $c = \frac{-6 \pm \sqrt{36 + 8}}{2}$ $c = \frac{-6 \pm \sqrt{44}}{2}$ $c = \frac{-6 \pm \sqrt{4 \cdot 11}}{2}$ $c = \frac{-6 \pm 2\sqrt{11}}{2}$ $c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $0$
Find Solutions: Determine the two solutions for $c$ . The two solutions are: $c = f = -3 + \sqrt{11}$ $c = g = -3 - \sqrt{11}$ Since f > g, we choose the solution with the positive square root.
Choose Correct Answer: Choose the correct answer from the given options. $\newline$ The value of $f$ is $-3 + \sqrt{11}$ , which matches option $(B)$ .

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