# Thread: Solving for x with square root

1. ## Solving for x with square root

x^2 + 2x + a = -(1/2) + sqroot(1/4 - a)

How do I solve for x?

Trying to get the x on the same side, so x(x + 2) = -(1/2) + sqroot(1/4 - a) - a

But can not get any further

2. Square both sides of the equation.

3. x^2 + 2x + a = -(1/2) + sqroot(1/4 - a)

x^2 +2x = -a -1/2 +sqrt(1/4 -a)
x^2 +2x +1 = 1 -a -1/2 +sqrt(1/4 -a)
(x +1)^2 = 1/2 -a +sqrt(1/4 -a)

I am not successful yet on how to make the RHS a perfect square too.
So, for now,

x +1 = +,-sqrt[1/2 -a +sqrt(1/4 -a)]
x = -1 +,-sqrt[1/2 -a +sqrt(1/4 -a)] ------------answer.

4. Originally Posted by Tikitac
x^2 + 2x + a = -(1/2) + sqroot(1/4 - a)

How do I solve for x?

Trying to get the x on the same side, so x(x + 2) = -(1/2) + sqroot(1/4 - a) - a

But can not get any further
This is the same as ticbol's answer, but in a different way.

a is a constant, so:
$\displaystyle x^2 + 2x + a = -\frac{1}{2} + \sqrt{ \frac{1}{4} - a}$

$\displaystyle x^2 + 2x + \left ( \frac{1}{2} + a - \sqrt{ \frac{1}{4} - a} \right ) = 0$

This is of the form $\displaystyle bx^2 + cx + d = 0$, so we may use the quadratic formula:
$\displaystyle x = \frac{ -c \pm \sqrt{c^2 - 4bd} }{2b}$
(Sorry for the shift in variables, but you already have an "a" in your expression. I didn't want to confuse the symbols.)

So
$\displaystyle x = \frac{-2 \pm \sqrt{4 - 4 \left ( \frac{1}{2} + a - \sqrt{ \frac{1}{4} - a} \right ) }}{2}$

$\displaystyle x = \frac{-2 \pm 2\sqrt{1 - \frac{1}{2} - a + \sqrt{ \frac{1}{4} - a} }}{2}$

$\displaystyle x = -1 \pm \sqrt{ \frac{1}{2} - a + \sqrt{ \frac{1}{4} - a} }$

-Dan