How do you solve ${(20 - r)}^{\frac{1}{2}} = r$ $(20-r)^(1/2)=r$ ?

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Answer 1 · 2016-08-02T01:42:56

$r = 4 or - 5$ $r= 4 or -5$

${(20 - r)}^{\frac{1}{2}} = r$ $(20-r)^(1/2) = r$

$\sqrt{20 - r} = r$ $sqrt(20-r) = r$

Square both sides:
$20 - r = r^{2}$ $20-r=r^2$

$r^{2} + r - 20 = 0$ $r^2+r-20=0$

Factorise:
$(r - 4) (r + 5) = 0$ $(r-4)(r+5)=0$

$r = 4 or - 5$ $r = 4 or -5$

How do you solve (20-r)^(1/2)=r(20−r)12=r(20-r)^(1/2)=r?