This free Solving Square Root Equations Calculator tool evaluates the equation and finds the variable value easily. You can use this calculator easily by just giving the square root equation as the input and hit on the calculate button to obtain the result within no time.
Solving Square Root Equations Calculator: Do you want to learn the concept of solving the square root equation? If yes, then stay tuned to this page. In the below sections, you can find the lengthy manual procedure to solve any type of square equation easily. To get the immediate results, make use of the online Solving Square Root Equations Calculator tool. We are also giving an example problem which is helpful for the better understanding of concept.
Finding variable value from a square root equation is not that simple. It contains several mathematical operations. You are advised to follow the simple guidelines provided below while solving the square root equations.
Example
Question: Question: Solve √2x+9 = 5?
Solution:
Given equation is
√2x+9 = 5
Apply square on the both sides
(√2x+9)2 = 52
2x+9= 25
Subtract 9 on the both sides of equation
2x+9-9=25-9
2x=25-9
2x=16
Now, divide both sides of equation divide by 2
2x/2=16/2
x=8
Want to finish your assignment in maths subject by understanding the solution of each concept then go with our comprehensive array of calculators prevailing on the Onlinecalculator.guru portal.
1. What is meant by Square Root Equation?
Square root equation is an equation which is present inside the root symbol. Usually, the radical equation has an independent variable. It means the radical equation or square root equation is in the form √x=y.
2. How can you solve Square Root Equations using a calculator?
The following steps are helpful for solving the square root equations using calculator:
3. Mention some examples of square root equations?
Generally, the square root equation is in the form of √x=y. The examples of square root equations are √(2x−5) = 1 + √(x−1), √x+29 = 52, √x2+2x+2 = 5.
4. Solve√(2x−5) − √(x−1) = 1?
√(2x−5) − √(x−1) = 1
√(2x−5) =1+√(x−1)
Apply the square function on both sides of equation
[√(2x−5)]2 =[1+√(x−1)]2
2x-5 = 1+2√(x−1)+x-1
2x-5 = x+2√(x−1)
2x-x-5=2√(x−1)
x-5/2=√(x−1)
Again apply square function on two sides
(x-5/2)2=[√(x−1)]2
[x2-10x+25]/4=x-1
x2-10x+25=4x-4
x2-10x+25-4x+4=0
x2-14x+29=0
By solving the above quadratic equation, we get x as 2.53 and 11.47