What are the solutions of equation : square root (5x-4)+square root(11x+4)=4square rootx?

Expert Answers

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We have to solve sqrt (5x-4) + sqrt(11x+4) = 4*sqrt x

sqrt (5x-4) + sqrt(11x+4) = 4*sqrt x

square both the sides

=> (5x - 4) + (11x + 4) + 2*sqrt (5x-4)*sqrt(11x+4) = 16x

=> 16x + 2*sqrt (5x-4)*sqrt(11x+4) = 16x

=> 2*sqrt (5x-4)*sqrt(11x+4) = 0

=> (5x - 4)(11x + 4) = 0

5x - 4 = 0 , gives x = 4/5

11x + 4 = 0, gives x = -4/11. But sqrt (-4/11) is not real.

Therefore the real value of x is 4/5.

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