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

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Answer 1 · 2015-06-29T22:50:31

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

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

$y = x^2+2x+1 = (x+1)^2$

hence $x = -1+-sqrt(y)$

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

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

Suppose you have a quadratic:

$x^2 + 6x + 9$

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:

$x^2 + 6x + 9 = (x+3)^2$

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

Similarly:

$4x^2 - 4x + 1 = (2x-1)^2$ like $441 = 21^2$

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