How do you solve ${(2 r - 5)}^{2} = 81$ $(2r-5)^2=81$ ?

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How do you solve ${(2 r - 5)}^{2} = 81$ $(2r-5)^2=81$ ?

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Answer 1 · 2016-08-05T10:38:14

$r = 7$ $r=7$
$r = - 2$ $r=-2$

Explanation:

${(2 r - 5)}^{2} = 81$ $(2r-5)^2=81$
or
$2 r - 5 = \pm \sqrt{81}$ $2r-5=+-sqrt81$
or
$2 r - 5 = \pm 9$ $2r-5=+-9$
or
$2 r = 5 \pm 9$ $2r=5+-9$
or
$r = \frac{5 \pm 9}{2}$ $r=(5+-9)/2$
or
$r = \frac{5 + 9}{2}$ $r=(5+9)/2$
or
$r = \frac{14}{2}$ $r=14/2$
or
$r = 7$ $r=7$ -------------Ans $1$ $1$
or
$r = \frac{5 - 9}{2}$ $r=(5-9)/2$
or
$r = - \frac{4}{2}$ $r=-4/2$
or
$r = - 2$ $r=-2$ ------------Ans $2$ $2$

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How do you solve ${(2 r - 5)}^{2} = 81$ $(2r-5)^2=81$ ?

1 Answer

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How do you solve (2r-5)^2=81(2r−5)2=81(2r-5)^2=81?

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Explanation:

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How do you solve ${(2 r - 5)}^{2} = 81$ $(2r-5)^2=81$ ?