# quadratic equation w/absoulute values

• Jan 10th 2009, 05:53 PM
requal
quadratic equation w/absoulute values
how would you solve
$x^2+|2x-1|=3$
• Jan 10th 2009, 06:17 PM
skeeter
Quote:

Originally Posted by requal
how would you solve
$x^2+|2x-1|=3$

interesting ...

$|2x-1| = 3-x^2$

since $|2x-1| \geq 0$ , $-\sqrt{3} \leq x \leq \sqrt{3}$

case 1 ...

$2x-1 = 3-x^2$

$x^2+2x-4 = 0$

$x^2+ 2x + 1 - 5 = 0$

$(x + 1)^2 - 5 = 0$

$x+1 = \pm \sqrt{5}$

$x = -1 \pm \sqrt{5}$

one solution works ... $x = -1 + \sqrt{5}$

case 2 ...

$2x-1 = -(3 - x^2)$

$2x-1 = x^2 - 3$

$0 = x^2 - 2x - 2$

$0 = x^2 - 2x + 1 - 3$

$0 = (x - 1)^2 - 3$

$x-1 = \pm \sqrt{3}$

$x = 1 \pm \sqrt{3}$

one solution works ... $x = 1 - \sqrt{3}$
• Jan 10th 2009, 06:20 PM
Jhevon
Quote:

Originally Posted by requal
how would you solve
$x^2+|2x-1|=3$

recall, $|x| = \left \{ \begin{array}{lr} x & \text{ if } x \ge 0 \\ & \\-x & \text{ if } x < 0 \end{array} \right.$

thus,

$x^2 + |2x - 1| = 3$

$\Rightarrow |2x - 1| = 3 - x^2$

$\Rightarrow 2x - 1 = 3 - x^2$ ...........(if $2x - 1 \ge 0$)

or

$-(2x - 1) = 3 - x^2$ ...........(if $2x - 1 < 0$)

now solve each equation independently, and then take the interval that covers both solutions