unit 8 homework 7 solving quadratics by square roots

Solving Quadratics by Square Root Method

How to solve quadratic equations using the square root method.

This is the “best” method whenever the quadratic equation  only contains [latex]{x^2}[/latex] terms. That implies no presence of any [latex]x[/latex] term being raised to the first power somewhere in the equation.

The general approach is to collect all [latex]{x^2}[/latex] terms on one side of the equation while keeping the constants to the opposite side. After doing so, the next obvious step is to take the square roots of both sides to solve for the value of [latex]x[/latex]. Always attach the [latex] \pm [/latex] symbol when you get the square root of the constant.

Examples of How to Solve Quadratic Equations by Square Root Method

Example 1 : Solve the quadratic equation below using the Square Root Method.

I will isolate the only [latex]{x^2}[/latex] term on the left side by adding both sides by [latex] + 1[/latex]. Then solve the values of [latex]x[/latex] by taking the square roots of both sides of the equation. As I mentioned before, we need to attach the plus or minus symbol to the square root of the constant.

So I have [latex]x = 5[/latex] and [latex]x = – \,5[/latex] as final answers since both of these values satisfy the original quadratic equation. I will leave it to you to verify.

Example 2 : Solve the quadratic equation below using the Square Root Method.

This problem is very similar to the previous example. The only difference is that after I have separated the [latex]{x^2}[/latex] term and the constant in the opposite sides of the equation, I need to divide the equation by the coefficient of the squared term before taking the square roots of both sides.

The final answers are [latex]x = 4[/latex] and [latex]x = – \,4[/latex].

Example 3 : Solve the quadratic equation below using the Square Root Method.

I can see that I have two [latex]{x^2}[/latex] terms, one on each side of the equation. My approach is to collect all the squared terms of [latex]x[/latex] to the left side, and combine all the constants to the right side. Then solve for [latex]x[/latex] as usual, just like in Examples 1 and 2.

The solutions to this quadratic formula are [latex]x = 3[/latex] and [latex]x = – \,3[/latex].

Example 4 : Solve the quadratic equation below using the Square Root Method.

The two parentheses should not bother you at all. The fact remains that all variables come in the squared form, which is what we want. This problem is perfectly solvable using the square root method.

So my first step is to eliminate both of the parentheses by applying the distributive property of multiplication. Once they are gone, I can easily combine like terms. Keep the [latex]{x^2}[/latex] terms to the left, and constants to the right. Finally, apply square root operation in both sides and we’re done!

Not too bad, right?

Example 5 : Solve the quadratic equation below using the Square Root Method.

Since the [latex]x[/latex]-term is being raised to the second power twice, that means, I need to perform two square root operations in order to solve for [latex]x[/latex].

The first step is to have something like this: (   ) 2 = constant . This allows me to get rid of the exponent of the parenthesis on the first application of square root operation.

After doing so, what remains is the “stuff” inside the parenthesis which has an [latex]{x^2}[/latex] term. Well, this is great since I already know how to handle it just like the previous examples.

There’s an [latex]x[/latex]-squared term left after the first application of square root.

Now we have to break up [latex]{x^2} = \pm \,6 + 10[/latex] into two cases because of the “plus” or “minus” in [latex]6[/latex].

• Solve the first case where [latex]6[/latex] is positive .

• Solve the second case where [latex]6[/latex] is negative .

The solutions to this quadratic equations are [latex]x = 4[/latex], [latex]x = – \,4[/latex], [latex]x = 2[/latex], and [latex]x = – \,2[/latex]. Yes, we have four values of [latex]x[/latex] that can satisfy the original quadratic equation.

Example 6 : Solve the quadratic equation below using the Square Root Method.

Example 7 : Solve the quadratic equation below using the Square Root Method.

You might also like these tutorials:

• Solving Quadratic Equations by Factoring Method
• Solving Quadratic Equations by the Quadratic Formula
• Solving Quadratic Equations by Completing the Square

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Solving Quadratic Equations by Taking Square Roots

Factoring Roots Completing the Square Formula Graphing Examples

Let's take another look at that last problem on the previous page:

Solve x 2 − 4 = 0

On the previous page, I'd solved this quadratic equation by factoring the difference of squares on the left-hand side of the equation, and then setting each factor equal to zero, etc, etc. The solution was " x  = ± 2 ". However—

I can also try isolating the squared-variable term on the left-hand side of the equation (that is, I can try getting the x 2 term by itself on one side of the "equals" sign), by moving the numerical part (that is, the 4 ) over to the right-hand side, like this:

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Solving by Taking Square Roots

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x 2 − 4 = 0

When I'm solving an equation, I know that I can do whatever I like to that equation as long as I do the exact same thing to both sides of that equation . On the left-hand side of this particular equation, I have an x 2 , and I want a plain old x . To turn the x 2 into an x , I can take the square root of each side of the equation, like this:

x = ± 2

Then the solution is x  = ±2 , just like it was when I solved by factoring the difference of squares.

Why did I need the " ± " (that is, the "plus-minus") sign on the 2 when I took the square root of the 4 ? Because I'm trying to find all values of the variable which make the original statement true, and it could have been either a positive 2 or a negative 2 that was squared to get that 4 in the original equation.

This duality is similar to how I'd had two factors, one "plus" and one "minus", when I used the difference-of-squares formula to solve this same equation on the previous page.

"Finding the solution to an equation" is a very different process from "evaluating the square root of a number". When finding "the" square root of a number, we're dealing exclusively with a positive value. Why? Because that is how the square root of a number is defined . The value of the square root of a number can only be positive, because that's how "the square root of a number" is defined.

Solving an equation, on the other hand — that is, finding all of the possible values of the variable that could work in an equation — is different from simply evaluating an expression that is already defined as having only one value.

Keep these two straight! A square-rooted number has only one value, but a square-rooted equation has two, because of the variable .

In mathematics, we need to be able to get the same answer, no matter which valid method we happen to have used in order to arrive at that answer. So, comparing the answer I got above with the answer I got one the previous page confirms that we must use the " ± " when taking square roots to solve.

(You may be doubting my work above in the step where I took the square root of either side, because I put a " ± " sign on only one side of the equation. Shouldn't I add this character to both sides of the equation? Kind of, yes. But if I'd put it on both sides of the equation, would anything really have changed? No. Try all the cases, if you're not sure.)

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

Solve x 2 − 50 = 0 .

This quadratic has a squared part and a numerical 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:

x 2 − 50 = 0

Then my 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 for solving some quadratic equations, but additional sorts of techniques allow us to find solutions to additional sorts of equations.

Solve ( x − 5) 2 − 100 = 0 .

This quadratic has a squared part and a numerical part. I'll start by adding the strictly-numerical term to the right-hand side of the equation, so that the squared binomial expression, containing the variable, is by itself on the left-hand side. Then I'll square-root both sides, remembering the " ± " on the numerical side, and then I'll 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

This equation, after taking the square root of either side, did not contain any radcials. Because of this, I was able to simplify my results, all the way down to simple values. My answer is:

x = −5, 15

The previous equation is an example of a equation where the careless student will omit the " ± " while solving, and will then have no clue as to how the book got the answer " x  = −5, 15 ".

These students get in the bad habit of not bothering to write the " ± " sign until they check their answers in the back of the book and suddenly "remember" that they "meant" to put the " ± " in there when they'd taken the square root of either side of the equation.

But this "magic" only works when you have the answer in the back (to remind you) and when the solution contains radicals (which doesn't always happen). In other cases, there will be no "reminder". Especially on tests, making the mistake of omitting the " ± " can be deadly. Don't be that student. Always remember to insert the " ± ".

By the way, since the solution to the previous equation consisted of integers, this quadratic could also have been solved by multiplying out the square, factoring, etc:

x 2 − 10 x + 25 − 100 = 0

x 2 − 10 x − 75 = 0

( x − 15)( x + 5) = 0

x − 15 = 0, x + 5 = 0

x = 15, −5

Solve ( x − 2) 2 − 12 = 0

This quadratic has a squared part and a numerical part. I'll add the numerical part over to the other side, so the squared part with the variable is by itself. Then I'll square-root both sides, remembering to add a " ± " to the numerical side, and then I'll simplify:

( x − 2) 2 − 12 = 0

( x − 2) 2 = 12

I can't simplify this any more. My answer is going to have radicals in it. My solution is:

This quadratic equation, unlike the one before it, could not have also been 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 part)" form? This concern leads to the next topic: solving by completing the square.

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This Quadratic Equations Unit Bundle contains guided notes, homework assignments, four quizzes, study guide and a unit test that cover the following topics:

• Introduction to Quadratic Equations (Standard Form, Vertex, Axis of Symmetry, Maximum, Minimum)

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• Vertex Form and Transformations

• Quadratic Roots (Identifying by Graphing)

• The Discriminant

• Solving Quadratic Equations by Factoring

• Solving Quadratic Equations by Square Roots

• Solving Quadratic Equations by Completing the Square

• Solving Quadratic Equations by the Quadratic Formula

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Unit 8 – Quadratic Functions

Introduction to Quadratic Functions

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The Leading Coefficient of a Quadratic

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Solving Quadratics by Completing the Square

Area and Completing the Square

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Solving Linear-Quadratic Systems

Quadratic Word Problems

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