Equations

Algebra Level 2

$\begin{cases} y = x^2 \\ y = 14 + \sqrt{x} \end{cases}$

If real numbers $x$ and $y$ satisfy the above relations, find the value of $x$ .

The answer is 4.

1 solution

Aman Dubey
Apr 10, 2016

Above equation can be expressed as $x^2 = 14 + \sqrt x$ Put $t = \sqrt x$ Thus we get, $t^4 = 14 + t$ $\Rightarrow t (t^3 -1) = 14$ As 14 can be expressed in product of two numbers as $2\times 7$ So $t$ must be 2 Which implies that $x = 4$

Equations

The answer is 4.

1 solution

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