# How do you solve x^2 - 4x +13 = 0 ?

Nov 4, 2015

This equation has no solutions

#### Explanation:

To solve this equation, rather than the classic formula, completing the square may come in handy.

Completing the square means that we can try to find a binomial square "hidden" in the equation and isolate it, and then deal with the rest.

The formula for the square of a binomial is the following:

${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$.

So, we need two squares, and a third terms, which is twice the multiplication of the bases of the squares.

Your equation starts with ${x}^{2} - 4 x$. Of course, ${x}^{2}$ is the square of $x$, so we wanto $- 4 x$ to be twice the multiplication of $x$ and another number. This other number is obviously $- 2$, so we can conclude that

${\left(x - 2\right)}^{2} = {x}^{2} - 4 x + 4$.

Your equation differs this expression for a difference of $9$ unit, in fact

${x}^{2} - 4 x + 13 = \left({x}^{2} - 4 x + 4\right) + 9$

From this point, we are able to tell that the equation has no solution: we want

$\left({x}^{2} - 4 x + 4\right) + 9 = {\left(x - 2\right)}^{2} + 9$

to be zero, but since a square is always positive, how can a sum of two positive quantities be zero? In other terms, you would have

${\left(x - 2\right)}^{2} + 9 = 0 \iff {\left(x - 2\right)}^{2} = - 9$

and again, a square can't be equal to a negative number.