How do you solve the equation #abs(x^2-1)=3#?

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Answer 1 · 2017-04-24T07:18:38

If this is a quadratic equation then #x=-2 or x=2#

#x^2-1=3#

#x^2-1-3=3-3#

#x^2-4=0#

This is a quadratic equation, in which ax^2+bx+c=0.

To solve for x the equation is #x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}#

So #x=\frac{-0+\sqrt{0^2-4\cdot \1\(-4\)}}{2\cdot \1}:\quad 2#

And #x=\frac{-0-\sqrt{0^2-4\cdot \1\(-4\)}}{2\cdot \1}:\quad -2#

So #x=-2 or 2#

The dot means multiplication.