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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, squared, plus, b, x, plus, c, equals, 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, squared, minus, 7, equals, 5 and asked to solve for x.
We can show our work like this:
\begin{aligned} 3x^2-7&=5\\\\ 3x^2&=12\\\\ x^2&=4\\\\ \sqrt{x^2}&=\pm \sqrt{4}\\\\ x&=\pm 2 \end{aligned}
So our two solutions are:
• x, equals, 2
• x, equals, minus, 2
Notice the plus minus 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, equals, 2x, equals, minus, 2
\begin{aligned}3x^2-7&=5\\\\3(2)^2-7&=5\\\\3\cdot4-7&=5\\\\12-7&=5\\\\5&=5\end{aligned}\begin{aligned}3x^2-7&=5\\\\3(-2)^2-7&=5\\\\3\cdot4-7&=5\\\\12-7&=5\\\\5&=5\end{aligned}
Yes! Both solutions check out.

### Example 2

We're given left parenthesis, x, minus, 3, right parenthesis, squared, minus, 81, equals, 0 and asked to solve for x.
We can show our work like this:
\begin{aligned} (x - 3)^2 - 81 &= 0\\\\ (x - 3)^2 &= 81\\\\ \sqrt{(x - 3)^2} &= \pm \sqrt{81}\\\\ x - 3 &= \pm 9\\\\ x &= \pm 9+3 \end{aligned}
So our two solutions are:
• x, equals, plus, 9, plus, 3, equals, start color #11accd, 12, end color #11accd
• x, equals, minus, 9, plus, 3, equals, start color #11accd, minus, 6, end color #11accd
Let's check both solutions:
x, equals, start color #11accd, 12, end color #11accdx, equals, start color #11accd, minus, 6, end color #11accd
\begin{aligned}(x - 3)^2 - 81 &= 0\\\\(\blueD{12} - 3)^2 - 81 &= 0\\\\9^2 - 81 &= 0\\\\81 - 81 &= 0\\\\0 &= 0\end{aligned}\begin{aligned}(x - 3)^2 - 81 &= 0\\\\(\blueD{-6} - 3)^2 - 81 &= 0\\\\(-9)^2 - 81 &= 0\\\\81 - 81 &= 0\\\\0 &= 0\end{aligned}
Yep! Both check out.
Practice
Solve for x.
left parenthesis, x, plus, 1, right parenthesis, squared, minus, 36, equals, 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!
(1 vote)
• 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.
• I think I understand this.
• how do you solve 2x(squared)-9x+19
(1 vote)
• 1) You have an expression, not an equation. You can only "solve" equations.
2) If you meant it to be: 2x^2 - 9x + 19 = 0
Then, it can be solved by either completing the square or by the quadratic formula.
• Do all quadratics always have two solutions?
(1 vote)
• No, not all quadratics have multiple solutions. In fact, some quadratic equations have no solutions. Here is how to tell how many solutions a quadratic equation has.

Suppose we have a quadratic equation a*x^2+b*x+c=0.

If b^2-4*a*c > 0, there are two solutions.
If b^2-4*a*c = 0, there is one solution.
If b^2-4*a*c < 0, there are no solutions.
• 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 can i get a better understanding
(1 vote)
• did I just accidentally do the wrong course?
(1 vote)
• can you use fancy quadratics skills for this
(1 vote)
• You can but it would take more time and be inefficient.
(1 vote)
• 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.