How to solve for x using the quadratic equation when there are multiple variables?

Question

Can someone please help me solve for $x$ using the quadratic formula for this question? I am not sure how to set this up.

$x^2+2ax-4b^2$

Answer 1

$$\displaystyle x=\frac{-2a\pm\sqrt{4a^2+4(4b^2)}}{2}$$ $$=-a\pm\frac{\sqrt{4a^2+16b^2}}{2}$$ $$=-a\pm\sqrt{a^2+4b^2}$$

Answer 2

Just as all quadratic equations: $a$ and $b$ are parameters and there may by be conditions to be satisfied by these parameters for real roots to exist. You can rewrite this equation as $$x^2+2ax-4b^2=(x+a)^2-a^2-4b^2$$ so the equation becomes $$(x+a)^2=a^2+4b^2\iff x=-a\pm\sqrt{a^2+4b^2}.$$

Answer 3

Firstly, remember that a quadratic equation always has the form:

$$y=ax^2 + bx + c$$

An example is $y=3x^2 + 4x + 5$

Since your equation already has $a,b$, I'm going to change them to $p,q$ so we don't get confused.

So your equation is $y=x^2+2px-4q^2$

So, just line them up:

In your case we have $y=x^2+2px-4q^2$

Thus, our $a=1, b=2p, c=-4q^2$

Therefore, we know our quadratic equation is:

$$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}=\dfrac{-(2p)\pm\sqrt{(2p)^2-4(1)(-4q^2)}}{2(1)}$$

Simplifying a little more, we get the following:

$$x=\dfrac{-(2p)\pm\sqrt{(2p)^2-4(1)(-4q^2)}}{2(1)}=-p\pm \dfrac{\sqrt{4(p^2+4q^2)}}{2}=-p\pm \sqrt{p^2+4q^2}$$

Now we can plug in what you had back as $p,q$ to obtain:

$$x=-a\pm\sqrt{a^2+4b^2}$$