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.

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