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

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Answer 1 · 2015-07-20T23:31:25

Let $t = sqrt(x)$ , solve the resulting quadratic in $t$ ,

then derive $x=4$ or $x=9$ .

Let $t=sqrt(x)$ .

Note $t >= 0$ since $sqrt$ denotes the non-negative square root.

Then $5t - t^2 = 6$

Add $t^2-5t$ to both sides to get:

$0 = t^2-5t+6 = (t-2)(t-3)$

So $t=2$ or $t=3$ . These both satisfy $t >= 0$ so are valid solutions.

So $x = 2^2 = 4$ or $x = 3^2 = 9$

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