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#