value of x

\hspace{-16}$Find value of $\mathbf{x}$ in $\mathbf{\sqrt{2x^2-4x+4}+\sqrt{2x^2-12x+26}=\sqrt{26}}$

2 Answers

sqrt(2 x^2-12 x+26)+sqrt(2 x^2-4 x+4) = sqrt(26)

sqrt(2 x^2-12 x+26) = sqrt(26)-sqrt(2 x^2-4 x+4)
Square both sides:
2 x^2-12 x+26 = (sqrt(26)-sqrt(2 x^2-4 x+4))^2

2 x^2-12 x+26 = 2 x^2-2 sqrt(26) sqrt(2 x^2-4 x+4)-4 x+30

2 sqrt(26) sqrt(2 x^2-4 x+4)-8 x-4 = 0

2 sqrt(26) sqrt(2 x^2-4 x+4) = 8 x+4

sqrt(2 x^2-4 x+4) = (8 x+4)/(2 sqrt(26))
Square both sides:
2 x^2-4 x+4 = 1/104 (8 x+4)^2

2 x^2-4 x+4 = (8 x^2)/13+(8 x)/13+2/13

(18 x^2)/13-(60 x)/13+50/13 = 0

(x-5/3)^2 = 0

x-5/3 = 0

x = 5/3

1708

man111 singh ·2011-10-18 10:05:30

yes hemang right answer

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