Print-friendly page

Solving Quadratic Equations:
     Solving by Taking Square Roots (page 2 of 6)

Sections: Solving by: factoring, taking roots, completing the square, using the formula, graphing

Let's take another look at that last problem:

• Solve x² – 4 = 0.

When you have "squared part minus a number", you can put the number over on the other side, like this:

x² – 4 = 0
x² = 4

Remember that, when solving an equation, you can do whatever you like to it, as long as you do the same thing to both sides. On the left-hand side, I have x², and I need just x. To turn an x² into an x, I square-root both sides:



x = ± 2

Then the solution is x = ± 2.

Why the "±" ("plus-or-minus") sign? Because it might have been a positive 2 or a negative 2 that was squared to get the 4.

(This solution is therefore different from "Evaluate the square root of 4", where you know you're dealing with a positive result. Solving an equation —finding all the values that work— is different from just evaluating an expression that is defined to have only one value. Remember: a square root has one value, but a square-rooted equation has two, because of the variable.)

This answer with the "±" matches the solution I got when I solved this equation by factoring, using the Difference of Squares formula. So this confirms the need to use the "±" when solving by square-rooting. And this square-rooting process allows us to solve some quadratics that we could not solve before. For instance:

• Solve x² – 50 = 0.

This quadratic has a squared part and a number part. Add the number to the other side (so the squared part is by itself), and square-root both sides. Remember to simplify:

x² – 50 = 0
x² = 50


Then the solution is   .

While we could have gotten the previous solution by factoring, we could never have gotten this solution by factoring. So factoring is useful, but other techniques allow us to find more solutions.

• Solve (x – 5)² – 100 = 0.

This quadratic has a squared part and a number part. Add the number to the other side (so the squared part is by itself), and square-root both sides. Remember to simplify:

(x – 5)² – 100 = 0
(x – 5)² = 100

x – 5 = ±10
x = 5 ± 10
x = 5 – 10  or  x = 5 + 10
x = –5   or  x = 15

Since the equation, after square-rooting, did not contain any square roots, I was able to simplify down to simple values. Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

The solution is x = –5, 15.

This example points out the importance of remembering the "±" when you square-root both sides. Many students develop a habit of not bothering to write the "±" sign until they check their answers in the back of the book and "remember" that they "meant" to put the "±" in there. However, that generally only works when the solution involves square roots (and only when you actually have the solution, unlike on tests, where you don't). This is an example of a problem where the careless student will omit the "±" and then have no clue how the book got the answer "x = – 5, 10". Don't be that student; always remember to insert the "±".

By the way, this quadratic could also have been solved by multiplying out the square and simplifying to get "x² – 10x – 75", which factors as "(x – 15)(x + 5)".

• Solve (x – 2)² – 12 = 0

This quadratic has a squared part and a number part. Add the number to the other side (so the squared part is by itself), and square-root both sides. Remember to simplify:

(x – 2)² – 12 = 0
(x – 2)² = 12


Then the solution is   .

This quadratic, unlike the previous one, could not be solved by factoring. We had to be given the quadratic in square-rootable form. This leads to the next topic: "completing the square".

<< Previous  Top  |  1 | 2 | 3 | 4 | 5 | 6  |  Return to Index  Next >>

  Copyright © 2006-2008  Elizabeth Stapel   |   About   |   Terms of Use

 Feedback   |   Error?