How do you solve #sqrt(10–x)+x=8#?

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Answer 1 · 2016-03-05T08:25:38

#x=9,6#

#sqrt(10-x)+x=8#

#rarrsqrt(10-x)=8-x#

Square both sides in order to get rid of the radical sign

#rarr(sqrt(10-x))^2=(8-x)^2#

Use the formula #(a-b)^2=a^2-2ab+b^2#

#rarr10-x=64-16x+x^2#

#rarr10=64-16x+x^2+x#

#rarr10=64-15x+x^2#

#rarr0=64-15x+x^2-10#

#rarr0=54-15x+x^2#

Rewrite

#rarrx^2-15x+54=0#

Factor the equation

#rarr(x-9)(x-6)=0#

If we solve for it we get #x=9,6#

But,

If we take the value of #9# of #x# into the equation,

#sqrt(10-9)+9=0#

#sqrt1+9=0#

#1+9!=0#

We should take #sqrt1# as #-sqrt1#

Hope! I had given the right answer!