How do you solve $25 v^{2} = 1$ $25v^2=1$ ?

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How do you solve $25 v^{2} = 1$ $25v^2=1$ ?

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Answer 1 · 2016-09-04T14:26:11

$v = \pm \frac{1}{5}$ $v=+-1/5$

Explanation:

Divide both sides by 25 giving:

$v^{2} = \frac{1}{25}$ $v^2=1/25$

Thus:

$\sqrt{v^{2}} = \sqrt{\frac{1}{25}}$ $sqrt(v^2)=sqrt(1/25)$

$v = \frac{\sqrt{1}}{\sqrt{25}} = \frac{1}{\sqrt{25}}$ $v=sqrt(1)/(sqrt(25)) = 1/sqrt(25)$

But as this is square root we should write $= \pm \frac{1}{\sqrt{25}}$ $=+-1/sqrt(25)$

$v = \pm \frac{1}{5}$ $v=+-1/5$

Answer 2 · 2016-09-04T14:29:34

$v = \pm \frac{1}{5}$ $v=+-1/5$

Explanation:

Begin by isolating $v^{2}$ $v^2$ . To do this we divide both sides of the equation by 25.

$rArr(cancel(25)^1 v^2)/cancel(25)^1=1/25rArrv^2=1/25$

now take the $color(blue)"square root of both sides"$

$rArrsqrt(v^2)=+-sqrt(1/25)$

$rArrv=+-1/5$

Answer 3 · 2016-09-04T16:42:42

$v = +-1/5$

Explanation:

Apart from the method shown by other contributors, we can also follow the usual method for a quadratic equation.

$rarr "Make it equal to 0 " rarr " find the factors."$

$25v^2 = 1$

$25v^2 -1 = 0 larr" difference of squares"$

$(5v+1)(5v-1) = 0$

Either factor could be 0

$5v+1 = 0 " " rarr v = -1/5$

$5v-1 = 0 " " rarr v = 1/5$

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