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Simple quadratic equations like x^2=4 can be solved by taking the square root. This article reviews several examples and gives you a chance to practice on your own.
In general, a quadratic equation can be written as:
$a{x}^{2}+bx+c=0$
In this article, we review how to solve quadratics that are solvable by taking the square root—no fancy factoring or quadratic equations here; we'll get to that technique later.

### Example 1

We're given $3{x}^{2}-7=5$ and asked to solve for $x$.
We can show our work like this:
$\begin{array}{rl}3{x}^{2}-7& =5\\ \\ 3{x}^{2}& =12\\ \\ {x}^{2}& =4\\ \\ \sqrt{{x}^{2}}& =±\sqrt{4}\\ \\ x& =±2\end{array}$
So our two solutions are:
• $x=2$
• $x=-2$
Notice the $±$ symbol we included when taking the square root of both sides. This symbol means "plus or minus," and it is important because it ensures we catch both solutions. Want a deeper explanation? Check out this video.
Let's check both solutions:
$x=2$$x=-2$
$\begin{array}{rl}3{x}^{2}-7& =5\\ \\ 3\left(2{\right)}^{2}-7& =5\\ \\ 3\cdot 4-7& =5\\ \\ 12-7& =5\\ \\ 5& =5\end{array}$$\begin{array}{rl}3{x}^{2}-7& =5\\ \\ 3\left(-2{\right)}^{2}-7& =5\\ \\ 3\cdot 4-7& =5\\ \\ 12-7& =5\\ \\ 5& =5\end{array}$
Yes! Both solutions check out.

### Example 2

We're given $\left(x-3{\right)}^{2}-81=0$ and asked to solve for $x$.
We can show our work like this:
$\begin{array}{rl}\left(x-3{\right)}^{2}-81& =0\\ \\ \left(x-3{\right)}^{2}& =81\\ \\ \sqrt{\left(x-3{\right)}^{2}}& =±\sqrt{81}\\ \\ x-3& =±9\\ \\ x& =±9+3\end{array}$
So our two solutions are:
• $x=+9+3=12$
• $x=-9+3=-6$
Let's check both solutions:
$x=12$$x=-6$
$\begin{array}{rl}\left(x-3{\right)}^{2}-81& =0\\ \\ \left(12-3{\right)}^{2}-81& =0\\ \\ {9}^{2}-81& =0\\ \\ 81-81& =0\\ \\ 0& =0\end{array}$$\begin{array}{rl}\left(x-3{\right)}^{2}-81& =0\\ \\ \left(-6-3{\right)}^{2}-81& =0\\ \\ \left(-9{\right)}^{2}-81& =0\\ \\ 81-81& =0\\ \\ 0& =0\end{array}$
Yep! Both check out.
Practice
Solve for $x$.
$\left(x+1{\right)}^{2}-36=0$

Want more practice? Check out this exercise

## Want to join the conversation?

• For the question, I got answer choice b?
Is that correct, because even though I got it correct, I feel like it is wrong. How can I check my answer? PLEASE HELP!
• Let's say x + 6 = square root of 10. How would I do this if 10 isn't a square number?
• This is a common scenario. Your results will have 2 terms.
First, let's fix an error in what you've done so far. Quadratic equations generally have 2 solutions. You lost one of them.
When you apply the square root to both sides of the equation, on the side with the 10, you need to ensure you have both the positive and negative root.
x+6 = +/- sqrt(10)

From here, you just need to subtract 6 from both sides of the equation.
x = -6 +/- sqrt(10)

Or, written individually, x = -6+sqrt(10) and x=-6-sqrt(10).

Should you need to use them for graphing, you can do estimated values for the square roots, then subtract 6.

Hope this helps.
• how do you solve (x-7)^2-9=49?
by extracting square root
• Add 9 to both sides of the equation: (x-7)^2=58
Take square root of both sides: sqrt(x-7)^2 = +/- sqrt(58)
Simplify square roots: x-7 = +/- sqrt(58)
Add 7 to both sides: x = 7 +/- sqrt(58)

Hope this helps.
• how can i get a better understanding
• did I just accidentally do the wrong course?
• how to solve : -(-x+1)^2
• Solving this is impossible without an equation. If they want you to simplify, remember PEMDAS. Parenthesis, Exponents, Multiplication and Division, Addition and Subtraction.

In this case, you can't really do anything in the Parenthesis, so you start with the exponent. First find what (-x+1) * (-x+1) is equal to. Then multiply everything by -1 to deal with the minus side at the beginning.

If you are trying to solve the equation (or find the "0"s), set the equation equal to 0. From that point, you would treat it like anything else that looks like this. Multiply both sides by -1 to get:

(-x+1)^2 = 0

From there, you should be able to use the review above to find your answer.
• How do I solve x^2-6x +8 = 0 by taking the square root? I know the solution is 2,4, but how do I use this method to solve?
• x^2-6x+8=0
As shown above you need to first simply the equation then put it in the quadratic form:
x = -(-6)±√(-6)^2-4*8 / 2

Then you square the -6 inside the parenthesis:
x = -(-6)±√36-4*8 / 2

Then multiply the -4 * 8:
x = -(-6)±√36-32 / 2

Subtract 36 and 32:
x = -(-6)±√4 / 2

Therefore take the square root of 4:
x = -(-6)±2 / 2

Opposite of -6 is 6:
x = 6±2 / 2

Then solve the equation for ± to be addition; 6+2:
8/2

x = 4

Ok now solve equation when ± is subtraction; 6-2:
x = 4/2

x = 2

solutions: 2 and 4 by using the quadratic formula.

Hopes this helps you out.
• When I was young I studied this equation (x + 3)^2 in a different way, by multiplying the terms by themselves.
Is there any refresh of this topic here?
I don't remember its name, but anyway, thank you for your time and good job with this software, I LOVE IT.
(1 vote)
• This is really helpful; thanks for creating an exercise like this to further increase my understanding of quadratic equations. I am only in Grade 7, and thanks to these carefully laid-out instructions and helpful tips, I am able to understand this concept. But I do have one question. If you have a standard three degree equation like "x^2 + 7x + 10," there are only three degrees. Why is it called a QUADratic equation? Can anybody help me out here?
(1 vote)
• SO … WHY THE NAME QUADRATIC? Quadratic equations are intimately connected with problems about squares and quadrangles (another name for rectangles). In fact, the word quadratic is derived from the Latin word quadratus for square. Square being represented by the ^2 degree. Hope this helps and keep up the good work 7th grader!
(1 vote)
• Create a list of steps, in order, that will solve the following equation.
$(x-5)^2=25$
(1 vote)
• I'm assuming your problem is: (x-5)^2=25 and that the extra brackets and back slashes have no meaning (not sure why they are there).

Look at example 2 on this page. Your steps would be similar to those found in that problem starting from: (x-3)^2=81
(1 vote)