Square Root Property: Definition with Solved Examples & FAQs
Beny Basumatary
| Updated: Dec 2, 2022 17:09 IST
Square root property can be used to easily solve quadratic equations that do not contain a linear term. This property becomes really helpful when you need to save time, solving quadratics. In this article, you will learn how to solve quadratic equations (without linear term) using the square root properties. We have covered the step-by-step procedure along with some solved examples on square root property as well.
What is Square Root Property?
The square root property is yet another easy method to solve quadratic equations. However, you can only solve a quadratic equation without a linear term using this property, i.e., an equation of the form \(ax^2=c \, or \,(ax+b)^2=c, \, where \,a\neq0\), can be easily solved using the square root property.
If \(x^2=k\) , where k is any constant number, then x= \(x= \pm\sqrt{k} \,(and\, k\neq0)\)
Hence, \(x= +\sqrt{k} \,and\, x= -\sqrt{k}\).
How To Use Square Root Property?
We can use square root property to solve quadratic equations of the form \(ax^2=c \,or\, (ax+b)^2=c\) by following the steps given below:
Step 1: The first step is to bring all the terms having a squared variable (\(x^2\)) to one side of the equation.
Step 2: Take the square root on both sides of the equation.
Step 3: Simplify to find values of x.
Learn Square Root of 169.
Square Root Property Examples
Following are some important square root property examples of quadratic equations in one variable.
Example 1. Solve \(2x^2=8\) quadratic equation in one variable using the square root properties.
Solution: \(x^2=\frac{8}{2}=4\)
Taking square root on both sides
x= 4
x= \(\pm 2\)
Hence, x= \(+2\) , x= \(-2\).
Example 2. Solve \(15x^2-12=10x^2\) quadratic equation in one variable using the square root properties.
Solution: Bring all the \(x^2\) terms on one side
\(15x^2-10x^2=12\)
\(5x^2=12\)
\(x^2=\frac{12}{5}\)
Taking square root on both sides
x= \(\pm\frac{12}{5}\)
x= \(\pm\frac{2\sqrt{3}}{\sqrt{5}}\)
Hence, x= \(+\frac{2\sqrt{3}}{\sqrt{5}}\) , x= \(-\frac{2\sqrt{3}}{\sqrt{5}}\)
Example 3. Solve \(3x^2=240\) quadratic equation in one variable using the square root properties.
Solution: \(x^2=\frac{240}{3}=80\)
Taking square root on both sides
x= \(\pm\sqrt{80}\)
x= \(\pm 4\sqrt{5}\)
Hence, x= \(+4\sqrt{5}\) , x= \(-4\sqrt{5}\).
Example 4. Solve \(x^2+7=5x^2\) quadratic equation in one variable using the square root properties.
Solution: Bring all the \(x^2\) terms on one side
\(x^2-5x^2= -7\)
\(-4x^2= -7\)
\(x^2=\frac{7}{4}\)
Taking square root on both sides
x= \(\pm\frac{\sqrt{7}}{\sqrt{4}}\)
x= \(\pm\frac{\sqrt{7}}{2}\)
Hence, x= \(+\frac{\sqrt{7}}{2}\), x= \(-\frac{\sqrt{7}}{2}\).
Example 5. Solve \(9x^2+5=53\) quadratic equation in one variable using the square root properties.
Solution: \(9x^2=53-5\)
\(9x^2=48\)
\(x^2=\frac{48}{9}\)
Taking square root on both sides
x= \(\frac{\sqrt{48}}{9}\)
x= \(\pm\frac{4\sqrt{3}}{3}\)
Hence, x= \(+\frac{4\sqrt{3}}{3}\), x= \(-\frac{4\sqrt{3}}{3}\).
Hopefully, this article was informative and useful for you. We believe, by now all your doubts regarding square root properties must be clear. If you still are doubtful about anything regarding square root properties or quadratic equations or algebra in general, you can contact us for the same. For more such questions and practice papers, you can download the Testbook App, for absolutely free. With this app, you can give mock tests and boost your preparations for any competitive exam.
| If you are checking Square Root Property article, also check the related maths articles: | |
| Square Root of 1 | How to find Square Root of a number |
| Square Root of Complex Number | Finding Square Root by Division Method |
| Square Root of 10 | Square Root of Decimals |
Square Root Property FAQs
Q.1 What is square root property?
Ans.1 According to the square root property If \(x^2=k\) , where k is any constant number, then
x = \(\pm\sqrt{k} \, where \,k\neq0\).
Q.2 What is the use of square root property?
Ans.2 Square root properties are yet another easy methods to solve quadratic equations. However, you can only solve a quadratic equation without a linear term using this property, i.e., an equation of the form \(ax^2=c \,or\, (ax+b)^2=c \,where \,a\neq0\),can be easily solved using the square root property.
Q.3 How many roots do we get using the square root properties?
Ans.3 You get 2 roots using the square root properties.
Q.4 What is the nature of roots we get using square root properties?
Ans.4 If \(x^2=k\), where k is any constant real number, then
x=\(\pm\sqrt{k} \,where \,k\neq0\). If k is a positive real number, then x has two real roots, while if k is any negative real number then x has two imaginary roots.
Q.5 How to solve an equation with two x squared terms using square root properties?
Ans.5 If an equation has more than one x-square term then bring all of them together on one side of the equals sign. Add or subtract according to the question and then apply square root properties. Read the article above for such solved examples.
