Solving Quadratics by Taking the Square Root

Solving Quadratics by Taking the Square Root

Sometimes, you'll have an equation that has only two terms when simplified: a term with a variable squared and a constant. Here are some examples:

• x² + 3 = 0

• 17y² = -3

• 2g² + g - 13 = g + 14

• Wait a minute, does that last one really qualify? Yes! Be sure to simplify first. Now, let's look at the steps for solving these problems.

There are 4 simple steps for solving these problems:

• Simplify the equations by combining like terms.

• Make sure your quadratic term is on one side of the equation, while your constant is on the other side.

• Divide both sides by the coefficient of the quadratic term.

• Take the square root of both sides.

• **NOTE** When you see a square root symbol in math, it's assumed to be the positive root. When you solve a problem and actually take the square root of both sides, you need to consider both the positive and negative roots as answers.

Let's try an example: x² = 9

• It's already simplified.

• The quadratic term is on the left and the constant is on the right.

• The coefficient of x² is 1, so we don't need to divide both sides.

• Now, we take the square root of both sides, remembering both the positive AND negative roots:

•  $x=\pm\sqrt{9}$  

•  $x=\pm3$  or  $x=\left\{-3,\ 3\right\}$  

Now for a trickier example:

 $3x^2+2x-x-x^2-x+3=35$  

• Simplify:  $2x^2+3=35$  

• Subtract 3 from both sides:  $2x^2=32$  

• Divide both sides by 2:  $x^2=16$  

• Take the square root:  $x=\pm4$   or  $x=\left\{-4,\ 4\right\}$  

• The next step is for you to check your answer by substituting -4 or 4 in the original equation to verify.

Now, it's your turn.

Try out the next few problems yourself.

Multiple Choice

 $x^2-81=0$  

•  $x=9$  

•  $x=8$  
  

•  $x=40.5$  

•  $x=9\ and\ -9$  

Multiple Choice

 $2x^2-26=0$  

•  $x=\pm\sqrt{26}$  

•  $x=\frac{\pm\sqrt{26}}{2}$  

•  $x=\left\{-3.61,3.61\right\}$  

•  $x=\sqrt{13}$  

Multiple Select

 $f\left(x\right)=x^2-144$  

Find the roots of the function. Select all that apply.

Hint: To find the roots of a function, replace "f(x)" with "0" and solve. 

• -72
  

• -12

• 0

• 12

• 72

Poll

What do you think you should do if the constant is a negative? 

For example:  $x^2=-25$  

• Only use the negative answer. In this case, the answer is -5.

• Since you use the positive and negative roots, it doesn't matter. The answer is still  $\pm5$  .

• It's impossible to take the square root of a negative number. There's no real answer.

You cannot take the square root of a negative number without getting an imaginary result (yes, imaginary numbers do exist). If you see a question like this on practice or an evaluation, the answer is "no real solutions".

Imaginary numbers exist in the fourth dimension (seriously - this is true), where the unicorns live (this is maybe not so true).

That's it!

You've finished "basic training" for solving quadratics by taking the square roots. I have one thing to say: