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# Solving quadratics by taking square roots

Learn how to solve quadratic equations like x^2=36 or (x-2)^2=49.

#### What you will learn in this lesson

So far you have solved linear equations, which include constant terms—plain numbers—and terms with the variable raised to the first power, ${x}^{1}=x$.
You will now learn how to solve quadratic equations, which include terms where the variable is raised to the second power, ${x}^{2}$.
Here are a few examples of the types of quadratic equations you will learn to solve:
${x}^{2}=36$
$\left(x-2{\right)}^{2}=49$
$2{x}^{2}+3=131$
Now let's get down to business.

## Solving ${x}^{2}=36$‍  and similar equations

Suppose we want to solve the equation ${x}^{2}=36$. Let's first verbalize what the equation is asking us to find. It is asking us which number, when multiplied by itself, equals 36.
If this question sounds familiar to you, it's because this is the definition of the square root of 36, which is expressed mathematically as $\sqrt{36}$.
Now, this is how the complete solution of the equation looks:
$\begin{array}{rl}{x}^{2}& =36\\ \\ \sqrt{{x}^{2}}& =\sqrt{36}& & \text{Take the square root.}\\ \\ x& =±\sqrt{36}\\ \\ x& =±6\end{array}$
Let's review what went on in this solution.

### What the $±$‍  sign means

Note that every positive number has two square roots: a positive square root and a negative square root. For example, both $6$ and $-6$, when squared, equal $36$. Therefore, this equation has two solutions.
The $±$ is just an efficient way of representing this concept mathematically. For example, $±6$ means "either $6$ or $-6$".

### A note about inverse operations

When we solved linear equations, we isolated the variable by using inverse operations: If the variable had $3$ added to it, we subtracted $3$ from both sides. If the variable was multiplied by $4$, we divided both sides by $4$.
The inverse operation of taking the square is taking the square root. However, unlike the other operations, when we take the square root we must remember to take both the positive and the negative square roots.
Now solve a few similar equations on your own.
Problem 1
Solve ${x}^{2}=16$.
$x=±$

Problem 2
Solve ${x}^{2}=81$.
$x=±$

Problem 3
Solve ${x}^{2}=5$.
Choose 1 answer:

## Solving $\left(x-2{\right)}^{2}=49$‍  and similar equations

Here is how the solution of the equation $\left(x-2{\right)}^{2}=49$ goes:
$\begin{array}{rl}\left(x-2{\right)}^{2}& =49\\ \\ \sqrt{\left(x-2{\right)}^{2}}& =\sqrt{49}& & \text{Take the square root.}\\ \\ x-2& =±7\\ \\ x& =±7+2& & \text{Add 2.}\end{array}$
Therefore, the solutions are $x=9$ and $x=-5$.
Let's review what went on in this solution.

### Isolating $x$‍

Using the inverse operation of taking the square root, we removed the square sign. This was important in order to isolate $x$, but we still had to add $2$ in the last step in order to really isolate $x$.

### Understanding the solutions

Our work ended with $x=±7+2$. How should we understand that expression? Remember that $±7$ means "either $+7$ or $-7$." Therefore, we should split our answer according to the two cases: either $x=7+2$ or $x=-7+2$.
This gives us the two solutions $x=9$ and $x=-5$.
Now solve a few similar equations on your own.
Problem 4
Solve $\left(x+3{\right)}^{2}=25$.
Choose 1 answer:

Problem 5
Solve $\left(2x-1{\right)}^{2}=9$.
Choose 1 answer:

Problem 6
Solve $\left(x-5{\right)}^{2}=7$.
Choose 1 answer:

### Why we shouldn't expand the parentheses

Let's go back to our example equation, $\left(x-2{\right)}^{2}=49$. Suppose we wanted to expand the parentheses there. After all, this is what we do in linear equations, right?
Expanding the parentheses results in the following equation:
${x}^{2}-4x+4=49$
If we wanted to take the square root in this equation, we would have to take the square root of the expression ${x}^{2}-4x+4$, but it's not clear if $\sqrt{{x}^{2}-4x+4}$ can be rewritten as a nice expression.
In contrast, taking the square roots of expressions like ${x}^{2}$ or $\left(x-2{\right)}^{2}$ gives us nice expressions like $x$ or $\left(x-2\right)$.
Therefore, it's actually helpful in quadratic equations to keep things factored, because this allows us to take the square root.

## Solving $2{x}^{2}+3=131$‍  and similar equations

Not all quadratic equations are solved by immediately taking the square root. Sometimes we have to isolate the squared term before taking its root.
For example, to solve the equation $2{x}^{2}+3=131$ we should first isolate ${x}^{2}$. We do this exactly as we would isolate the $x$ term in a linear equation.
$\begin{array}{rl}2{x}^{2}+3& =131\\ \\ 2{x}^{2}& =128& & \text{Subtract 3.}\\ \\ {x}^{2}& =64& & \text{Divide by 2.}\\ \\ \sqrt{{x}^{2}}& =\sqrt{64}& & \text{Take the square root.}\\ \\ x& =±8\end{array}$
Now solve a few similar equations on your own.
Problem 7
Solve $3{x}^{2}-7=5$.
Choose 1 answer:

Problem 8
Solve $4\left(x-1{\right)}^{2}+2=38$.
Choose 1 answer:

Challenge problem
Solve ${x}^{2}+8x+16=9$.
Choose 1 answer:

## Want to join the conversation?

• what do you do if you have a hard time remembering big formulas?
(29 votes)
• Create random codes that will be easier to remember. For example "Q = I x t" is formula for charge used in a circuit for a given time. I call it "Quit" where Qu is Charge, i is Current and t is time.
Build your own logic that makes no sense. Australia has more letters than Europa, so it cannot fit in Europa. Australia is not located in Europa.
(77 votes)
• Wouldn't expanding the parenthesis make it easier? For example: (x-2)^2 = 81 be x^2- 2^2=81, thus making it easier to solve?
(17 votes)
• No! (x-2)² = (x-2)(x-2) = x²-2x-2x+4 = x²-4x-4
That means, now you gotta solve for x²-4x+4 = 81.
x²-4x+4 = 81x²-4x+4-81 = 81-81x²-4x-77 = 0x²-11x+7x-77 = 0 (splitting the middle term)
x(x-11)+7(x-11) = 0
(x+7)(x-11) = 0 (factoring out (x-11))
Using the zero product property,
x = -7 or x = 11
This way, as you can see, is much, much longer than the square root method. Why bother going the long way, when you can have the solution in just a few seconds?

Hope this helps! :)
(33 votes)
• when looking at the quadratic equation with the (x-#)^2
what if the ^2 was next to the (x) in the equation?
(16 votes)
• "Note that every positive number has two square roots: a positive square root and a negative square root. For example, both 6 and -6 when squared, equal 36. Therefore, this equation has two solutions."

In my calculator I tried squaring -6 but I got -36?
(9 votes)
• I know you posted this a while ago and may not see this but:

Your calculator (like mine) interprets "-6", as "-1 * 6".
Meaning it interprets "-6^2" as "-1 * 6^2", and since calculators obey the order of operations, it squares the 6 first, then multiplies that by -1.

If you want to square a negative number, then type "(-6)^2", which the calculator interprets as "(-1 * 6)^2", meaning it preforms the multiplication before squaring the number.

I hope that helped.
(20 votes)
• for the very last challenge question:

how does the equation go from x^2 + 8x + 16 --> (x + 4) ^2 ?

is it because 4*4 = 16 and 4+4=8 thus, (x+4)^2 ?
(10 votes)
• Yes. If you factor x^2 + 8x + 16 it becomes (x + 4)(x + 4), which is the same as (x + 4) ^2
(14 votes)
• Do you always work on the left side of the equal sign first in problems like the challenge question?
(8 votes)
• You don't need to... but it kind of goes over there later.
(5 votes)
• So I did most of the videos prior to this point but I still don't understand what a quadratic is or more specifically, how it differs from a polynomial.
(5 votes)
• A quadratic is a polynomial that (when simplified) can be written in the form: Ax^2 + Bx + C where A can not = 0. If it is a quadratic equation, then it would be: Ax^2 + Bx + C = 0. So what does all that mean... a quadratic is a polynomial that has 1, 2 or 3 terms, but the highest degree term will have a variable that is squared.
There are many polynomials that are not quadratics. For example: 3x + 7 = 0 is a polynomial equation. However, it can not be written in the form Ax^2 + Bx + C =0 because there is no "x^2" term. Thus, it is not a quadratic. Instead, 3x + 7 = 0 is a simple linear equation (or 1st degree equation) that can be solved without using quadratic methods
2nd example: x^3 + 5x^2 + 6 =0 is a 3rd degree polynomial equation, however it is not a quadratic because the highest degree term is x^3 (not x^2). You will learn that equations like this can sometimes be solved using a combination of quadratic methods (e.g., factoring is used to get down to a lower degree: X ( X^2 + 5X + 6) = 0. We now have 2 factors, where one is a quadratic and you could use an appropriate quadratic method to solve that factor).
Hope this helps.
(13 votes)
• What is the best way to tell the difference between the functions, and when to use which one?
(7 votes)
• Which functions, exactly? Please, be more specific and maybe I can help you!
(2 votes)
• In the challenge question, how do we go from x^2+8x+16=9 to (x+4)^2=9?
(4 votes)
• Factor the left side and you get: (x+4)(x+4) = (x+4)^2
Any time you complete the square, the trinomial is a perfect square trinomial. Its factors will be matching binomials. They always get written as a binomial squared in preparation for apply the square root method.
Hope this helps.
(8 votes)
• what happens if you cant remember the equations?
(4 votes)
• If you understand the process you should be able to go back and forth with ease.
(6 votes)