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## Algebra 1

### Course: Algebra 1 > Unit 14

Lesson 3: Solving by taking the square root- Solving quadratics by taking square roots
- Solving quadratics by taking square roots
- Quadratics by taking square roots (intro)
- Solving quadratics by taking square roots examples
- Quadratics by taking square roots
- Quadratics by taking square roots: strategy
- Quadratics by taking square roots: strategy
- Solving quadratics by taking square roots: with steps
- Quadratics by taking square roots: with steps
- Solving simple quadratics review

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# Solving simple quadratics review

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:

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:

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, 2 | x, 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:

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 #11accd | x, 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.

*Want to learn more about these types of problems? Check out this video.*

*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!(13 votes)- (x+1)^2 - 36

= (x+1)^2 = 36 | *Add 36 to both sides to cancel out.

= x+1 = +/-6 | *Take the plus-or-minus square root of both sides, since x could be positive or negative. 36 becomes plus-or-minus 6. If you would like a better understanding of this principle, watch this khan video. It explains it in detail starting at 1: 37.

https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:quadratic-functions-equations/x2f8bb11595b61c86:untitled-1082/v/simple-quadratic-equation

x = either (6-1 = 5) or (-6-1 = -7) | *Subtract 1 from both sides to cancel out. If 6 is positive, then 6 minus 1 is 5, but if 6 is negative, then -6 minus 1 is -7.

Hope this helps!(8 votes)

- how do you solve (x-7)^2-9=49?

by extracting square root(4 votes)- 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.(8 votes)

- Let's say x + 6 = square root of 10. How would I do this if 10 isn't a square number?(3 votes)
- 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).

Those are your solutions.

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

Hope this helps.(8 votes)

- how can i get a better understanding(3 votes)
- did I just accidentally do the wrong course?(3 votes)
- how to solve : -(-x+1)^2(3 votes)
- 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.(3 votes)

- 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?(3 votes)
- 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.(1 vote)

- Do all quadratics always have two solutions?(2 votes)
- 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.(2 votes)

- All these problems on zero product are making me nauseous.(2 votes)
- can you use fancy quadratics skills for this(1 vote)
- You can but it would take more time and be inefficient.(3 votes)