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(1)/(2)c^(2)+3c=1
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 ?
Choose 1 answer:
(A)
-3+sqrt2
(B)
-3+sqrt11
(c)
3+sqrt2
(D)
3+sqrt11

$\frac{1}{2} c^{2}+3 c=1$ $\newline$ 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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Solve multi-variable equations

Full solution

Q.

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

Set Equation to Zero: Given the quadratic equation $(1)/(2)c^{2} + 3c = 1$ , we first need to bring all terms to one side to set the equation to zero. $\newline$ $(1)/(2)c^{2} + 3c - 1 = 0$ $\newline$ Multiply through by $2$ to clear the fraction: $\newline$ $c^{2} + 6c - 2 = 0$
Use Quadratic Formula: Now we use the quadratic formula to solve for $c$ : $c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 1$ , $b = 6$ , and $c = -2$ .
Calculate Discriminant: Calculate the discriminant $(b^2 - 4ac)$ : $\text{Discriminant} = (6)^2 - 4(1)(-2) = 36 + 8 = 44$
Plug Values into Formula: Now plug the values into the quadratic formula: $\newline$ c = [ $-6 \pm \sqrt{44}$ ] / $2$ $\newline$ c = [ $-6 \pm \sqrt{4 \times 11}$ ] / $2$ $\newline$ c = [ $-6 \pm 2\sqrt{11}$ ] / $2$ $\newline$ c = $-3 \pm \sqrt{11}$
Find Solutions: We have two solutions for $c$ : $c = -3 + \sqrt{11}$ and $c = -3 - \sqrt{11}$ Since f > g, we choose the larger solution for $f$ .
Choose Larger Solution: The larger solution is $f = -3 + \sqrt{11}$ , which corresponds to answer choice (B).

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