$\sqrt{25 y^{2} + 40 y + 16} = 5 y + 4$ $sqrt(25y^2+40y+16) =5y+4$ Solve the equation?

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Answer 1 · 2017-04-06T16:52:22

It is proved that $\sqrt{{(25 y)}^{2} + 40 y + 16} = 5 y + 4$ $sqrt[(25y)^2+40y+16] = 5y+4$

Squaring both sides, we get

$\sqrt{25 y^{2} + 40 y + 16} = \sqrt{{(5 y)}^{2} + 2 \times 5 y \times 4 + {(4)}^{2}}$ $sqrt(25y^2+40y+16) = sqrt[(5y)^2+2xx5yxx4+(4)^2$

$\Rightarrow {[\sqrt{5 y + 4}]}^{2} = 5 y + 4$ $rArr [sqrt(5y+4)]^2 = 5y+4$ proved

√25y2+40y+16=5y+4sqrt(25y^2+40y+16) =5y+4 Solve the equation?