# How do you solve y = x^2 + 2x + 1?

Jun 29, 2015

To 'solve' $y = {x}^{2} + 2 x + 1$ could mean:

'How do you calculate $x$ for any given $y$?'

$y = {x}^{2} + 2 x + 1 = {\left(x + 1\right)}^{2}$

hence $x = - 1 \pm \sqrt{y}$

If $y = 0$ then $x = - 1$

#### Explanation:

Here's a little 'trick' to help spot some of these perfect square quadratics...

${x}^{2} + 6 x + 9$

The coefficients of the ${x}^{2}$, $x$ and constant terms are $1 , 6 , 9$

Does the pattern $1 , 6 , 9$ ring a bell?

$169 = {13}^{2}$

and we find:

${x}^{2} + 6 x + 9 = {\left(x + 3\right)}^{2}$

where $\left(x + 3\right)$ has terms with coefficients $1 , 3$

Similarly:

$4 {x}^{2} - 4 x + 1 = {\left(2 x - 1\right)}^{2}$ like $441 = {21}^{2}$