# 4.7.1 Solving "X Squared Equals Number"

The solution to the equation

x^2=a

where a≥0 is x=\pm\sqrt{a}

It seems like we've hit a brick wall. We can't factor the left-hand side of the equation below:

-4.9x^{2}+2.45x+1.7=0

so we need to try and develop a new technique. Welcome to the world of the mathematician! One door has just been closed and we hope that somewhere down the hall, there's another door that we can open. Let's see if we can develop a completely new approach to solving a quadratic equation that will work for us regardless of whether we can factor our quadratic.

As usual, let's start with a related question that's easier to solve; let's get rid of the 2.45x term for now. For example, let's try and solve:

-4.9x^2+1.7=0

Can you isolate the x? To three decimal places, we get that: x = or

You can see the full explanation below.

We can bring the 1.7 over to the right hand side to get that:

-4.9x^2=-1.7

We can then divide both sides by -4.9 to get that

x^2=\frac{-1.7}{-4.9}

Upon taking the square root of both sides (the opposite process of squaring), we see that this equation will have two solutions for x:

\sqrt{\frac{1.7}{4.9}}=.589\textrm{ or }-\sqrt{\frac{1.7}{4.9}}=-.589

If c is any positive number, the two solutions to x^2=c will be \pm \sqrt{c}.

Also, let's note that the equation solve x^{2}="negative number", has no solution since there's no number that we can square to get back a negative number), and the equation x^{2}=0 has only one solution, namely x=0.

Explore!

Solve x^2-9=0. Solution: x= or .

Now let's try an equation that's more interesting. For example, instead of x^2="number", let's try to solve (binomial)^2="number". With our binomial equal to x+2, solve (x+2)^2 =13 and then take a look at the full solution below.

Taking the square root of both sides and remembering that we need to consider both the positive and negative square roots, we have that (x+2)=\pm \sqrt{13} so that x = -2 \pm \sqrt{13}.

Explore!

Solve (x+4)^2=16

Solution: x = or .

Solve (2x+1)^2=49

Solution: x = or .

Moral of the story: If we can get our original equation to look like "something" squared = number, we'll be able to solve it. This is a very different path to take than our method of factoring: let's see where this new method takes us!