How many solutions does $1 x^{2} + 2 x + 1 = 0$ $1x^2+2x+1=0$ have?

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Answer 1 · 2016-11-14T15:02:12

One solution: $x = - 1$ $x = -1$

Factoring the quadratic equation gives:

$(x + 1) (x + 1) = 0$ $(x + 1)(x + 1) = 0$

Because each term in the factoring is the same ( $x + 1$ $x + 1$ ) this problem has only 1 solution - $x = - 1$ $x = -1$

How many solutions does 1x^2+2x+1=01x2+2x+1=01x^2+2x+1=0 have?