How do you solve #sqrt(56-r)=r#?

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Answer 1 · 2016-12-19T15:43:54

#r = 7#

Square both sides to get

#56 - r = r^2#

Then gather terms to get

#r^2 + r - 56 = 0.#

This is a simple quadratic, which can be solved by the quadratic formula, or, more simply, by factoring, into

#(r - 7)(r +8) = 0#

Setting each factor equal to zero in turn gives

#r - 7 = 0 implies r = 8" "# and #" "r + 8 = 0 implies r = -8#

To check,

#sqrt(56 - 7) = sqrt(49) = 7#

#sqrt(56 - (-8)) = sqrt(64) != -8 -># this means that #r=-8# is not a solution to the original equation.