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, $x=9,4$

How do you solve 5(sqrt x) - x =65(sqrt x) - x =6?