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(v+(1)/(5))^(2)-9=0 What is the sum of the solutions to the given equation? Choose 1 answer: (A) -(3)/(5) (B) -(2)/(5) (C) -(1)/(5) (D) 0

(v+15)2−9=0 \left(v+\frac{1}{5}\right)^{2}-9=0 \newlineWhat is the sum of the solutions to the given equation?\newlineChoose 11 answer:\newline(A) −35 -\frac{3}{5} \newline(B) −25 -\frac{2}{5} \newline(C) −15 -\frac{1}{5} \newline(D) 00

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Full solution

Q. (v+15)2−9=0 \left(v+\frac{1}{5}\right)^{2}-9=0 \newlineWhat is the sum of the solutions to the given equation?\newlineChoose 11 answer:\newline(A) −35 -\frac{3}{5} \newline(B) −25 -\frac{2}{5} \newline(C) −15 -\frac{1}{5} \newline(D) 00
  1. Isolate squared term: First, let's isolate the squared term by adding 99 to both sides of the equation.\newline(v+15)2−9+9=0+9(v + \frac{1}{5})^2 - 9 + 9 = 0 + 9\newline(v+15)2=9(v + \frac{1}{5})^2 = 9
  2. Take square root: Now, take the square root of both sides to solve for vv.(v+15)2=±9\sqrt{(v + \frac{1}{5})^2} = \pm\sqrt{9}v+15=±3v + \frac{1}{5} = \pm3
  3. Subtract to isolate vv: Next, subtract 15\frac{1}{5} from both sides to get vv by itself.\newlinev+15−15=±3−15v + \frac{1}{5} - \frac{1}{5} = \pm3 - \frac{1}{5}\newlinev=±3−15v = \pm3 - \frac{1}{5}
  4. Calculate possible values: Now, calculate the two possible values for vv.v=3−15v = 3 - \frac{1}{5} and v=−3−15v = -3 - \frac{1}{5}v=2.8v = 2.8 and v=−3.2v = -3.2

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