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How to Solve Square Root Problems
Last Updated: November 3, 2022 References
This article was co-authored by David Jia . David Jia is an Academic Tutor and the Founder of LA Math Tutoring, a private tutoring company based in Los Angeles, California. With over 10 years of teaching experience, David works with students of all ages and grades in various subjects, as well as college admissions counseling and test preparation for the SAT, ACT, ISEE, and more. After attaining a perfect 800 math score and a 690 English score on the SAT, David was awarded the Dickinson Scholarship from the University of Miami, where he graduated with a Bachelor’s degree in Business Administration. Additionally, David has worked as an instructor for online videos for textbook companies such as Larson Texts, Big Ideas Learning, and Big Ideas Math. There are 11 references cited in this article, which can be found at the bottom of the page. This article has been viewed 334,866 times.
While the intimidating sight of a square root symbol may make the mathematically-challenged cringe, square root problems are not as hard to solve as they may first seem. Simple square root problems can often be solved as easily as basic multiplication and division problems. More complex square root problems, on the other hand, can require some work, but with the right approach, even these can be easy. Start practicing square root problems today to learn this radical new math skill!
Understanding Squares and Square Roots

- Try squaring a few more numbers on your own to test this concept out. Remember, squaring a number is just multiplying it by itself. You can even do this for negative numbers. If you do, the answer will always be positive. For example, (-8) 2 = -8 × -8 = 64 .

- As another example, let's find the square root of 25 (√(25)). This means we want to find the number that squares to make 25. Since 5 2 = 5 × 5 = 25, we can say that √(25) = 5 .
- You can also think of this as "undoing" a square. For example, if we want to find √(64), the square root of 64, let's start by thinking of 64 as 8 2 . Since a square root symbol basically "cancels out" a square, we can say that √(64) = √(8 2 ) = 8 .

- On the other hand, numbers that don't give whole numbers when you take their square roots are called imperfect squares . When you take one of these numbers' square roots, you usually get a decimal or fraction. Sometimes, the decimals involved can be quite messy. For instance, √(13) = 3.605551275464...

- 1 2 = 1 × 1 = 1
- 2 2 = 2 × 2 = 4
- 3 2 = 3 × 3 = 9
- 4 2 = 4 × 4 = 16
- 5 2 = 5 × 5 = 25
- 6 2 = 6 × 6 = 36
- 7 2 = 7 × 7 = 49
- 8 2 = 8 × 8 = 64
- 9 2 = 9 × 9 = 81
- 10 2 = 10 × 10 = 100
- 11 2 = 11 × 11 = 121
- 12 2 = 12 × 12 = 144

- Let's say that we want to find the square root of 900. At first glance, this looks very difficult! However, it's not hard if we separate 900 into its factors. Factors are the numbers that can multiply together to make another number. For instance, since you can make 6 by multiplying 1 × 6 and 2 × 3, the factors of 6 are 1, 2, 3, and 6.
- Instead of working with the number 900, which is somewhat awkward, let's instead write 900 as 9 × 100. Now, since 9, which is a perfect square, is separated from 100, we can take its square root on its own. √(9 × 100) = √(9) × √(100) = 3 × √(100). In other words, √(900) = 3√(100) .
- We can even simplify this two steps further by dividing 100 into the factors 25 and 4. √(100) = √(25 × 4) = √(25) × √(4) = 5 × 2 = 10. So, we can say that √(900) = 3(10) = 30 .

- Note that although imaginary numbers can't be represented with ordinary digits, they can still be treated like ordinary numbers in many ways. For instance, the square roots of negative numbers can be squared to give those negative numbers, just like any other square root. For example, i 2 = -1
Using Long Division-Style Algorithms

- Start by writing out your square root problem in the same from as a long division problem. For example, let's say that we want to find the square root of 6.45, which is definitely not a convenient perfect square. First, we'd write an ordinary radical symbol (√), then we'd write our number underneath it. Next, we'd make a line above our number so that it's in a little "box" — just like in long division. When we're done, we should have a long-tailed "√" symbol with 6.45 written under it.
- We'll be writing numbers above our problem, so be sure to leave space.

- In our example, we would divide 6.45 into pairs like this: 6-.45-00 . Note that there is a "leftover" digit on the left — this is OK.

- In our example, the first group in 6-.45-00 is 6. The biggest number that is less than or equal to 6 when squared is 2 — 2 2 = 4. Write a "2" above the 6 under the radical.

- In our example, we would start by taking the double of 2, the first digit of our answer. 2 × 2 = 4. Next, we would subtract 4 from 6 (our first "group"), getting 2 as our answer. Next, we would drop down the next group (45) to get 245. Finally, we would write 4 once more to the left, leaving a small space to add onto the end, like this: 4_.

- In our example, we want to find the number to fill in the blank in 4_ × _ that makes the answer as large as possible but still less than or equal to 245. In this case, the answer is 5 . 45 × 5 = 225, while 46 × 6 = 276.

- Continuing from our example, we would subtract 225 from 245 to get 20. Next, we would drop down the next pair of digits, 00, to make 2000. Doubling the numbers above the radical sign, we get 25 × 2 = 50. Solving for the blank in 50_ × _ =/< 2,000, we get 3 . At this point, we have "253" above the radical sign — repeating this process once again, we get a 9 as our next digit.

- In our example, the number under the radical sign is 6.45, so we would simply slide the point up and place it between the 2 and 5 digits of our answer, giving us 2.539 .
Quickly Estimating Imperfect Squares

- For example, let's say we need to find the square root of 40. Since we've memorized our perfect squares, we can say that 40 is in between 6 2 and 7 2 , or 36 and 49. Since 40 is greater than 6 2 , its square root will be greater than 6, and since it is less than 7 2 , its square root will be less than 7. 40 is a little closer to 36 than it is to 49, so the answer will probably be a little closer to 6. In the next few steps, we'll narrow our answer down.

- In our example problem, a reasonable estimate for the square root of 40 might be 6.4 , since we know from above that the answer is probably a little closer to 6 than it is to 7.

- Multiply 6.4 by itself to get 6.4 × 6.4 = 40.96 , which is slightly higher than original number.
- Next, since we over-shot our answer, we'll multiply the number one tenth less than our estimate above by itself and to get 6.3 × 6.3 = 39.69 . This is slightly lower than our original number. This means that the square root of 40 is somewhere between 6.3 and 6.4 . Additionally, since 39.69 is closer to 40 than 40.96, you know the square root will be closer to 6.3 than 6.4.

- In our example, let's pick 6.33 for our two-decimal point estimate. Multiply 6.33 by itself to get 6.33 × 6.33 = 40.0689. Since this is slightly above our original number, we'll try a slightly lower number, like 6.32. 6.32 × 6.32 = 39.9424. This is slightly below our original number, so we know that the exact square root is between 6.33 and 6.32 . If we wanted to continue, we would keep using this same approach to get an answer that's continually more and more accurate.
Expert Q&A Did you know you can get expert answers for this article? Unlock expert answers by supporting wikiHow

Support wikiHow by unlocking this expert answer.

- For quick solutions, use a calculator. Most modern calculators can instantly find square roots. Usually, all you need to do is to simply type in your number, then press the button with the square root symbol. To find the square root of 841, for example, you might press: 8, 4, 1, (√) and get an answer of 29 . [16] X Research source ⧼thumbs_response⧽ Helpful 0 Not Helpful 0

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- ↑ David Jia. Academic Tutor. Expert Interview. 14 January 2021.
- ↑ https://virtualnerd.com/algebra-foundations/powers-square-roots/powers-exponents/squaring-a-number
- ↑ https://www.mathsisfun.com/square-root.html
- ↑ http://virtualnerd.com/algebra-1/algebra-foundations/powers-square-roots/square-roots/square-root-estimation
- ↑ https://www.cuemath.com/algebra/perfect-squares/
- ↑ https://www.khanacademy.org/math/algebra/rational-exponents-and-radicals/alg1-simplify-square-roots/a/simplifying-square-roots-review
- ↑ https://math.libretexts.org/Courses/Monroe_Community_College/MTH_165_College_Algebra_MTH_175_Precalculus/00%3A_Preliminary_Topics_for_College_Algebra/0.03%3A_Review_-_Radicals_(Square_Roots)
- ↑ http://www.homeschoolmath.net/teaching/square-root-algorithm.php
- ↑ https://www.cuemath.com/algebra/square-root-by-long-division-method/
- ↑ https://virtualnerd.com/algebra-1/algebra-foundations/powers-square-roots/square-roots/square-root-estimation
- ↑ http://www.math.com/students/calculators/source/square-root.htm
About This Article

To solve square root problems, understand that you are finding the number that, when multiplied by itself, equals the number in the square root. For quick recall, memorize the first 10-12 perfect squares, so that you recognize the square root of numbers like 9, 25, 49, or 121. If possible, break the number under the square root into individual perfect squares. For example, √(900) can be broken into √(9) × √(100), and √(100) can be broken into √(25) × √(4), reducing the problem to √(9) × √(25) × √(4), or 3 x 5 x 2 for an answer of 30. If you want to learn how to estimate imperfect square roots, keep reading the article! Did this summary help you? Yes No
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Unit 10: Lesson 2
- Intro to square-root equations & extraneous solutions
- Square-root equations intro
- Intro to solving square-root equations
Solving square-root equations
- Solving square-root equations: one solution
- Solving square-root equations: two solutions
- Solving square-root equations: no solution
Introduction
Practice question 1: isolating the radical term.
- Your answer should be
- an integer, like 6 6 6 6
- a simplified proper fraction, like 3 / 5 3/5 3 / 5 3, slash, 5
- a simplified improper fraction, like 7 / 4 7/4 7 / 4 7, slash, 4
- a mixed number, like 1 3 / 4 1\ 3/4 1 3 / 4 1, space, 3, slash, 4
- an exact decimal, like 0.75 0.75 0 . 7 5 0, point, 75
- a multiple of pi, like 12 pi 12\ \text{pi} 1 2 pi 12, space, start text, p, i, end text or 2 / 3 pi 2/3\ \text{pi} 2 / 3 pi 2, slash, 3, space, start text, p, i, end text
Practice question 2: Two possible solutions
- (Choice A) x = 0 x=0 x = 0 x, equals, 0 A x = 0 x=0 x = 0 x, equals, 0
- (Choice B) x = − 4 x=-4 x = − 4 x, equals, minus, 4 B x = − 4 x=-4 x = − 4 x, equals, minus, 4
- (Choice C) x = 4 x=4 x = 4 x, equals, 4 C x = 4 x=4 x = 4 x, equals, 4
- (Choice D) x = − 6 x=-6 x = − 6 x, equals, minus, 6 D x = − 6 x=-6 x = − 6 x, equals, minus, 6
- (Choice E) x = 1 x=1 x = 1 x, equals, 1 E x = 1 x=1 x = 1 x, equals, 1
Practice question 3
- (Choice B) x = − 5 x=-5 x = − 5 x, equals, minus, 5 B x = − 5 x=-5 x = − 5 x, equals, minus, 5
- (Choice C) x = 1 x=1 x = 1 x, equals, 1 C x = 1 x=1 x = 1 x, equals, 1
- (Choice D) x = − 5 2 x=-\dfrac{5}{2} x = − 2 5 x, equals, minus, start fraction, 5, divided by, 2, end fraction D x = − 5 2 x=-\dfrac{5}{2} x = − 2 5 x, equals, minus, start fraction, 5, divided by, 2, end fraction
- (Choice E) None of the above E None of the above
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Online Calculators

Square Root Calculator

Calculator Use
Use this calculator to find the principal square root and roots of real numbers. Inputs for the radicand x can be positive or negative real numbers. The answer will also tell you if you entered a perfect square.
The answer will show you the complex or imaginary solutions for square roots of negative real numbers. See also the Simplify Radical Expressions Calculator to simplify radicals instead of finding fractional (decimal) answers.
Square Roots, odd and even:
There are 2 possible roots for any positive real number. A positive root and a negative root. Given a number x , the square root of x is a number a such that a 2 = x . Square roots is a specialized form of our common roots calculator .
"Note that any positive real number has two square roots, one positive and one negative. For example, the square roots of 9 are -3 and +3, since (-3) 2 = (+3) 2 = 9. Any nonnegative real number x has a unique nonnegative square root r; this is called the principal square root .......... For example, the principal square root of 9 is sqrt(9) = +3, while the other square root of 9 is -sqrt(9) = -3. In common usage, unless otherwise specified, "the" square root is generally taken to mean the principal square root."[1].
Perfect Square Calculator
This calculator will also tell you if the number you entered is a perfect square or is not a perfect square. A perfect square is a number x where the square root of x is a number a such that a 2 = x and a is an integer. For example, 4, 9 and 16 are perfect squares since their square roots, 2, 3 and 4, respectively, are integers.
Example Square Roots:
- The 2nd root of 81, or 81 radical 2, or the square root of 81 is written as $$ \sqrt[2]{81} = \sqrt[]{81} = \pm 9 $$.
- The 2nd root of 25, or 25 radical 2, or the square root of 25 is written as $$ \sqrt[2]{25} = \sqrt[]{25} = \pm 5 $$.
- The 2nd root of 100, or 100 radical 2, or the square root of 100 is written as $$ \sqrt[2]{100} = \sqrt[]{100} = \pm 10 $$.
- The 2nd root of 10, or 10 radical 2, or the square root of 10 is written as $$ \sqrt[2]{10} = \sqrt[]{10} = \pm 3.162278 $$.
To calculate fractional exponents use our calculator for Fractional Exponents .
[1] Weisstein, Eric W. "Square Root." From MathWorld -- A Wolfram Web Resource. Square Root
Square Roots Calculator
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9.6: Solve Equations with Square Roots
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Learning Objectives
By the end of this section, you will be able to:
- Solve radical equations
- Use square roots in applications
Before you get started, take this readiness quiz.
- Simplify: ⓐ \(\sqrt{9}\) ⓑ \(9^2\). If you missed this problem, review Example 9.1.1 and Exercise 1.3.22 .
- Solve: 5(x+1)−4=3(2x−7). If you missed this problem, review Exercise 2.4.16 .
- Solve: \(n^2−6n+8=0\). If you missed this problem, review Exercise 7.6.13 .
Solve Radical Equations
In this section we will solve equations that have the variable in the radicand of a square root. Equations of this type are called radical equations.
Definition: RADICAL EQUATION
An equation in which the variable is in the radicand of a square root is called a radical equation .
As usual, in solving these equations, what we do to one side of an equation we must do to the other side as well. Since squaring a quantity and taking a square root are ‘opposite’ operations, we will square both sides in order to remove the radical sign and solve for the variable inside.
But remember that when we write \(\sqrt{a}\) we mean the principal square root. So \(\sqrt{a} \ge 0\) always. When we solve radical equations by squaring both sides we may get an algebraic solution that would make \(\sqrt{a}\) negative. This algebraic solution would not be a solution to the original radical equation ; it is an extraneous solution. We saw extraneous solutions when we solved rational equations, too.
Example \(\PageIndex{1}\)
For the equation \(\sqrt{x+2}=x\):
- Is x=2 a solution?
- Is x=−1 a solution?
1. Is x=2 a solution?
2. Is x=−1 a solution?
Example \(\PageIndex{2}\)
For the equation \(\sqrt{x+6}=x\):
- Is x=−2 a solution?
- Is x=3 a solution?
Example \(\PageIndex{3}\)
For the equation \(\sqrt{−x+2}=x\):
- Is x=1 a solution?
For\(a \ge 0\), \((\sqrt{a})^2=a\)
How to Solve Radical Equations
Example \(\PageIndex{4}\)
Solve: \(\sqrt{2x−1}=7\)

Example \(\PageIndex{5}\)
Solve: \(\sqrt{3x−5}=5\).
Example \(\PageIndex{6}\)
Solve: \(\sqrt{4x+8}=6\).
Definition: SOLVE A RADICAL EQUATION.
- Isolate the radical on one side of the equation.
- Square both sides of the equation.
- Solve the new equation.
- Check the answer.
Example \(\PageIndex{7}\)
Solve: \(\sqrt{5n−4}−9=0\).
Example \(\PageIndex{8}\)
Solve: \(\sqrt{3m+2}−5=0\).
\(\frac{23}{3}\)

Example \(\PageIndex{9}\)
Solve: \(\sqrt{10z+1}−2=0\).
\(\frac{3}{10}\)
Example \(\PageIndex{10}\)
Solve: \(\sqrt{3y+5}+2=5\).
Example \(\PageIndex{11}\)
Solve: \(\sqrt{3p+3}+3=5\).
\(\frac{1}{3}\)
Example \(\PageIndex{12}\)
Solve: \(\sqrt{5q+1}+4=6\).
\(\frac{3}{5}\)
When we use a radical sign, we mean the principal or positive root. If an equation has a square root equal to a negative number, that equation will have no solution.
Example \(\PageIndex{13}\)
Solve: \(\sqrt{9k−2}+1=0\).
Example \(\PageIndex{14}\)
Solve: \(\sqrt{2r−3}+5=0\)
no solution
Example \(\PageIndex{15}\)
Solve: \(\sqrt{7s−3}+2=0\).
Definition: BINOMIAL SQUARES
\[\begin{array}{cc} {(a+b)^2=a^2+2ab+b^2}&{(a−b)^2=a^2−2ab+b^2}\\ \nonumber \end{array}\]
Don’t forget the middle term!
Example \(\PageIndex{16}\)
Solve: \(\sqrt{p−1}+1=p\).
Example \(\PageIndex{17}\)
Solve: \(\sqrt{x−2}+2=x\).
Example \(\PageIndex{18}\)
Solve: \(\sqrt{y−5}+5=y\).
Example \(\PageIndex{19}\)
Solve: \(\sqrt{r+4}−r+2=0\).
Example \(\PageIndex{20}\)
Solve: \(\sqrt{m+9}−m+3=0\).
Example \(\PageIndex{21}\)
Solve: \(\sqrt{n+1}−n+1=0\)
When there is a coefficient in front of the radical, we must square it, too.
Example \(\PageIndex{22}\)
Solve: \(3\sqrt{3x−5}−8=4\).
Example \(\PageIndex{23}\)
Solve: \(\sqrt{24a+2}−16=16\).
\(\frac{127}{2}\)
Example \(\PageIndex{24}\)
Solve: \(\sqrt{36b+3}−25=50\).
\(\frac{311}{3}\)
Example \(\PageIndex{25}\)
Solve: \(\sqrt{4z−3}=\sqrt{3z+2}\).
Example \(\PageIndex{26}\)
Solve: \(\sqrt{2x−5}=\sqrt{5x+3}\).
Example \(\PageIndex{27}\)
Solve: \(\sqrt{7y+1}=\sqrt{2y−5}\).
Sometimes after squaring both sides of an equation, we still have a variable inside a radical. When that happens, we repeat Step 1 and Step 2 of our procedure. We isolate the radical and square both sides of the equation again.
Example \(\PageIndex{28}\)
Solve: \(\sqrt{m}+1=\sqrt{m+9}\).
Example \(\PageIndex{29}\)
Solve: \(\sqrt{x}+3=\sqrt{x+5}\).
Example \(\PageIndex{30}\)
Solve: \(\sqrt{m}+5=\sqrt{m+16}\).
Example \(\PageIndex{31}\)
Solve: \(\sqrt{q−2}+3=\sqrt{4q+1}\).
Example \(\PageIndex{32}\)
Solve: \(\sqrt{y−3}+2=\sqrt{4y+2}\).
no solution
Example \(\PageIndex{33}\)
Solve: \(\sqrt{n−4}+5=\sqrt{3n+3}\).
Use Square Roots in Applications
As you progress through your college courses, you’ll encounter formulas that include square roots in many disciplines. We have already used formulas to solve geometry applications.
We will use our Problem Solving Strategy for Geometry Applications, with slight modifications, to give us a plan for solving applications with formulas from any discipline.
Definition: SOLVE APPLICATIONS WITH FORMULAS.
- Read the problem and make sure all the words and ideas are understood. When appropriate, draw a figure and label it with the given information.
- Identify what we are looking for.
- Name what we are looking for by choosing a variable to represent it.
- Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
- Solve the equation using good algebra techniques.
- Check the answer in the problem and make sure it makes sense.
- Answer the question with a complete sentence.
We used the formula A=L·W to find the area of a rectangle with length L and width W . A square is a rectangle in which the length and width are equal. If we let s be the length of a side of a square, the area of the square is \( s^2\) .

The formula \(A=s^2\) gives us the area of a square if we know the length of a side. What if we want to find the length of a side for a given area? Then we need to solve the equation for s .
\[\begin{array}{ll} {}&{A=s^2}\\ {\text{Take the square root of both sides.}}&{\sqrt{A}=\sqrt{s^2}}\\ {\text{Simplify.}}&{s=\sqrt{A}}\\ \nonumber \end{array}\]
We can use the formula \(s=\sqrt{A}\) to find the length of a side of a square for a given area.
Definition: AREA OF A SQUARE

We will show an example of this in the next example.
Example \(\PageIndex{34}\)
Mike and Lychelle want to make a square patio. They have enough concrete to pave an area of 200 square feet. Use the formula \(s=\sqrt{A}\) to find the length of each side of the patio. Round your answer to the nearest tenth of a foot.
Example \(\PageIndex{35}\)
Katie wants to plant a square lawn in her front yard. She has enough sod to cover an area of 370 square feet. Use the formula \(s=\sqrt{A}\) to find the length of each side of her lawn. Round your answer to the nearest tenth of a foot.
Example \(\PageIndex{36}\)
Sergio wants to make a square mosaic as an inlay for a table he is building. He has enough tile to cover an area of 2704 square centimeters. Use the formula \(s=\sqrt{A}\) to find the length of each side of his mosaic. Round your answer to the nearest tenth of a foot.
Another application of square roots has to do with gravity.
Definition: FALLING OBJECTS
On Earth, if an object is dropped from a height of hh feet, the time in seconds it will take to reach the ground is found by using the formula,
\(t=\frac{\sqrt{h}}{4}\)
For example, if an object is dropped from a height of 64 feet, we can find the time it takes to reach the ground by substituting h=64 into the formula.
It would take 2 seconds for an object dropped from a height of 64 feet to reach the ground.
Example \(\PageIndex{37}\)
Christy dropped her sunglasses from a bridge 400 feet above a river. Use the formula \(t=\frac{\sqrt{h}}{4}\) to find how many seconds it took for the sunglasses to reach the river.
Exercise \(\PageIndex{38}\)
A helicopter dropped a rescue package from a height of 1,296 feet. Use the formula \(t=\frac{\sqrt{h}}{4}\) to find how many seconds it took for the package to reach the ground.
Example \(\PageIndex{39}\)
A window washer dropped a squeegee from a platform 196 feet above the sidewalk Use the formula \(t=\frac{\sqrt{h}}{4}\) to find how many seconds it took for the squeegee to reach the sidewalk.
3.5 seconds
Police officers investigating car accidents measure the length of the skid marks on the pavement. Then they use square roots to determine the speed, in miles per hour, a car was going before applying the brakes.
Definition: SKID MARKS AND SPEED OF A CAR
If the length of the skid marks is d feet, then the speed, s , of the car before the brakes were applied can be found by using the formula,
\(s=\sqrt{24d}\)
Example \(\PageIndex{40}\)
After a car accident, the skid marks for one car measured 190 feet. Use the formula \(s=\sqrt{24d}\) to find the speed of the car before the brakes were applied. Round your answer to the nearest tenth.
Example \(\PageIndex{41}\)
An accident investigator measured the skid marks of the car. The length of the skid marks was 76 feet. Use the formula \(s=\sqrt{24d}\) to find the speed of the car before the brakes were applied. Round your answer to the nearest tenth.
Example \(\PageIndex{42}\)
The skid marks of a vehicle involved in an accident were 122 feet long. Use the formula \(s=\sqrt{24d}\) to find the speed of the vehicle before the brakes were applied. Round your answer to the nearest tenth.
Key Concepts
- Check the answer. Some solutions obtained may not work in the original equation.

- On Earth, if an object is dropped from a height of hh feet, the time in seconds it will take to reach the ground is found by using the formula\(t=\frac{\sqrt{h}}{4}\).
- If the length of the skid marks is d feet, then the speed, s , of the car before the brakes were applied can be found by using the formula \(s=\sqrt{24d}\).

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