# How do you describe the nature of the roots of the equation x^2=4x-1?

Mar 13, 2017

This quadratic has two distinct irrational real roots.

#### Explanation:

Given:

${x}^{2} = 4 x - 1$

Subtract $4 x - 4$ from both sides to get:

${x}^{2} - 4 x + 4 = 3$

That is:

${\left(x - 2\right)}^{2} = 3$

Hence:

$x - 2 = \pm \sqrt{3}$

So:

$x = 2 \pm \sqrt{3}$

So this quadratic equation has two distinct irrational real roots.

Note that instead of the full derivation of the roots, we could have examined the discriminant...

Given:

${x}^{2} = 4 x - 1$

Subtract $4 x - 1$ from both sides to get:

${x}^{2} - 4 x + 1 = 0$

This is in standard form:

$a {x}^{2} + b x + c = 0$

with $a = 1$, $b = - 4$ and $c = 1$.

This has discriminant $\Delta$, given by the formula:

$\Delta = {b}^{2} - 4 a c = {\left(- 4\right)}^{2} - 4 \left(1\right) \left(1\right) = 16 - 4 = 12 = {2}^{2} \cdot 3$

Since $\Delta > 0$ we can deduce that our quadratic has two distinct real roots.

Note also that $\Delta$ is not a perfect square, so those roots are irrational.