How do you solve $x^{2} + 10 = 11$ $x^2+10=11$ ?

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Answer 1 · 2015-10-15T01:54:13

$\pm 1$ $+-1$

Subtract $10$ $10$ form both sides

$x^{2} + 10 - 10 = 11 - 10$ $x^2+10-10=11-10$

$x^{2} = 1$ $x^2=1$

Take the square root of both sides

$\sqrt{x^{2}} = \pm \sqrt{1}$ $sqrt(x^2)=+-sqrt(1)$

$x = \pm 1$ $x=+-1$

Remember we have to have plus or minus $1$ $1$ because
$1^{2} = 1$ $1^2=1$ and ${(- 1)}^{2} = 1$ $(-1)^2=1$

How do you solve x2+10=11x^2+10=11?