Using the Square Root Property

Mini Lecture Video

Key Takeaways

The General Method

All quadratic equations can be written in the form

$$a(x-b)^2-c=0$$

Where \(a,b,c\in \mathbb{R}\).

When a quadratic equation is written in this form, one can solve this equation by:

• Adding \(c\) to both sides, then
• dividing both sides by \(a\), then
• take the square root of both sides (making sure to write \(\pm\) on the right-hand-side\), then
• setting up two equations, add \(b\) to both sides.

The overall idea is to solve this similar to how you’d solve a linear equation, trying to get the squared term by itself, then cancel the square with a square root, then get \(x\) by itself.

To illustrate, suppose we have

$$2(x-5)^2-8=0$$

To solve this, notice that the above implies

$$2(x-5)^2=8$$

$$(x-5)^2=\frac{8}{2}=4$$

$$(x-5)=\pm \sqrt{4}$$

$$x-5=2\ \ \ and\ \ \ x-5=-2$$

$$x=7\ \ \ and\ \ \ x=3$$

Complex Numbers and Solutions

Sometimes we will be solving equations such as

$$x^2+1=0.$$

Upon adding \(1\) to both sides and taking square roots, this gives

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

But, as we know, \(\sqrt{-1}\) doesn’t exist. So, mathematicians decided, essentially out of necessity, to “invent” a new number called \(i\) to represent the so-called square root of \(-1\), noting that \(i^2=-1\) by this definition.

So, in solving an equation such as \(x^2+1=0\), getting \(x=\pm \sqrt{-1}\) we instead write \(x=\pm i\), and say that the solutions to this equation are purely imaginary or just imaginary.

For other equations, such as \((x-2)^2+9=0\), things get a little more complex. The first few steps give us that \((x-2)^2=-9\), which implies \((x-2)=\pm \sqrt{-9}=\pm 3i\). Adding \(2\) to both sides gives us an answer of \(x=2\pm 3i\) or

$$x=2+3i\ \ \ or\ \ \ x=2-3i$$

The above are what are called complex numbers, because they consist of an imaginary part and a real part. In the case of \(2+3i\), the real part is \(2\) and the imaginary part is \(3\). real and imaginary parts of a complex number cannot be combined via addition or subtraction since they are not “like-terms.”

Getting only complex numbers as a solution to an equation means that the equation has no real solution, meaning there is no such real number \(x\) so that the equation is true… so essentially, we have to force into existence the solutions for \(x\) that we seek!