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:

Previously, I'd solved this by factoring the difference of squares, and solving each factor; the solution was "x = ± 2". However—

I can also try isolating the squared variable term, putting the number over on the other side, like this:

x2 – 4 = 0
x2 = 4

I know that, when solving an equation, I can do whatever I like to that equation as long as I do the same thing to both sides of the equation. On the left-hand side of this particular equation, I have an x2, and I need a plain x. To turn an x2 into an x, I can take the square root of each side of the equation:

x = ± 2

Then the solution is x = ± 2

Why did I need the "±" ("plus-minus") sign on the 2 when I took the square root of the 4? Because it might have been a positive 2 or a negative 2 that was squared to get that 4 in the original equation.

Warning: Finding the solution, above, is a very different process from "evaluating the square root of 4". When finding the square root of a number, you know that you're dealing with a positive result, because that is how the square root of a number is defined. Solving an equation, on the other hand — finding all of the possible values of the variable that will work in the equation — is different from just evaluating an expression that is already defined to have only one value.

Remember: a square-rooted number has only one value, but a square-rooted equation has two, because of the variable.

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The answer I got above, with the "±" sign, matches the solution I got when I solved this equation using the difference-of-squares factoring formula. Thus, this result confirms the need to use the "±" sign when solving by square-rooting. (In mathematics, you need to be able to get the same answer, no matter which valid method you happen to have used in order to arrive at that answer.)

A benefit of this square-rooting process is that it allows us to solve some quadratics that we could not have solved before. For instance:

• Solve x2 – 50 = 0.

This quadratic has a squared part and a number part. I'll start by adding the numerical term to the other side of the equaion (so the squared part is by itself), and then I'll square-root both sides. I'll need to remember to simplify the square root:

x2 – 50 = 0
x2 = 50

Then the solution is

While we could have gotten the previous integer solution by factoring, we could never have gotten this radical solution by factoring. Factoring is clearly useful, but additional techniques can allow us to find solutions to additional sorts of equations.

• Solve (x – 5)2 – 100 = 0.

This quadratic has a squared part and a number part. I'll start by adding the number to the other side of the equation, so the squared binomial is by itself. Then I'll square-root both sides, remembering to simplify my results:

(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 2002-2011 All Rights Reserved

The solution is x = –5, 15

This example points out the importance of remembering the "±" when you square-root both sides:

Warning: Many students develop the bad 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 when they took the square roots of either side of their equation. However, that generally only works when the solution involves square roots (and only when you actually have the solution, which isn't the case on tests). The above 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, 15". Don't be that student; always remember to insert the "±".

By the way, since the solution to the equation consisted of nice neat "whole" numbers, this quadratic could also have been solved by multiplying out the square and simplifying to get "x2 – 10x – 75", which then could have been factored as "(x – 15)(x + 5)".

• Solve (x – 2)2 – 12 = 0

This quadratic has a squared part and a number part. I'll add the number to the other side (so the squared part is by itself), and then I'll square-root both sides, remembering to simplify:

(x – 2)2 – 12 = 0
(x – 2)2 = 12

Then the solution is

This quadratic equation, unlike the previous one, could not be solved by factoring. But how would I have solved it, if they had not given me the quadratic already put into "(squared part) minus (a number)" form? This concern leads to the next topic: "completing the square".

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 Cite this article as: Stapel, Elizabeth. "Solving Quadratic Equations: Solving by Taking Roots." Purplemath. Available from     http://www.purplemath.com/modules/solvquad2.htm. Accessed [Date] [Month] 2016

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