How do you solve #sqrt(2x-1) - sqrt(x+7) = 0#?

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Answer 1 · 2016-01-24T05:40:30

#x=8#

#sqrt(2x-1) - sqrt(x+7) = 0#

#=>sqrt(2x-1) = sqrt(x+7)#

#=> 2x-1 = x+7#

#=> 2x-x = 7+1#

#=> x = 8#

Answer 2 · 2016-02-21T09:56:49

#x=8#

#color(blue)(sqrt(2x-1)-sqrt(x+7)=0#

Add #sqrt(x+7)# both sides

#rarrsqrt(2x-1)=sqrt(x+7)#

Square both sides to get rid of the radical sign

#rarr(sqrt(2x-1))^2=(sqrt(x+7))^2#

#rarr2x-1=x+7#

Subtract #x# both sides

#rarrx-1=7#

#color(green)(rArrx=7+1=8#

Check

#color(brown)(sqrt(2(8)-1)-sqrt(8+7)=0#

#color(brown)(sqrt(16-1)-sqrt(8+7)=0#

#color(brown)(sqrt15-sqrt15=0#

So,It is true