How do you solve #5(sqrt x) - x =6#?

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Answer 1 · 2016-02-02T12:36:57

#5(sqrtx)-x=6#

#rarr5(sqrtx)=6+x#

#rarrsqrtx=(6+x)/5#

Square both sides to remove the radical sign:

#rarr(sqrtx)^2=((6+x)/5)^2#

#rarrx=((6+x)/5)((6+x)/5)#

#rarrx=(36+6x+6x+x^2)/25#

#rarrx=(36+12x+x^2)/25#

Cross multiply:

#rarr25x=36+12x+x^2#

Subtract #12x# both sides:

#rarr13x=36+x^2#

Subtract #13x# both sides:

#rarr0=36+x^2-13x#

Write the equation in Standard form:

#rarrx^2-13x+36=0#

Luckily it Factors to:

#rarr(x-9)(x-4)=0#

So,#x=9,4#