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A "radical" equation means that equation in which at least one unknown variable is inside a radical, like a square root or cube root etc.

The "radical" in "radical equations" can be any root, whether a square root, a cube root, or some other root, but the radical must involve an unknown variable in its radical.

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## How to Solve Radical Equations?

That is, if the radical has an integer or any other real number inside it and its unknown variable is free from radical then the equation will not be considered as a radical equation.

General Method to solve radical equations:
A radical equation is solved by squaring or cubing (whichever radical sign is present in the equation) both the sides of the equation, not the individual terms of the equation.

In the end, we will have to check the solutions got because when solving radical equations, squaring or cubing makes some extra incorrect solutions, which does not hold true for the equations.

Steps for Solving: To solve any radical equations we need to follow the following steps:

1) Make independent the radical (or one of the radicals) to one side of the equal to sign.

2) If the radical is a square root, we will square each side of the equation (and if the radical has some other index then, we will put each side to that power which is equal to the index of the root.)

3) Now we will solve the resulting equation as usually solved.

4) In the end, we will check the answer(s) to avoid any invalid roots.

### Solved Examples

Question 1: Solve the equation, $\sqrt{x-1}=x-7$
Solution:

$\sqrt{x-1}=x-7$; squaring both the sides
$(\sqrt{(x-1)})^2=(x-7)^{2}$ this implies
x – 1 = (x – 7)(x – 7)
=> x - 1 = x2 – 14x + 49
=>  x2 – 15x + 50 = 0
=> (x – 5)(x – 10) = 0, this implies x = 5 or x = 10.

Now we will check, which one of the above two values are valid for the equation,
For x = 5, the above equation becomes 2 = -2, which is not possible.
For x = 10, the above equation becomes 3 = 3, which hold true.
Hence, x = 10 is the only solution.

Question 2: Solve the equation $\sqrt{x-2}=5$
Solution:

Since this equation is in the form "(square root) = (number)", so, squaring both sides, we will get $(\sqrt{x-2})^{2}=5^{2}$ or  x – 2 = 25 or x = 27 which is the only solution.

Question 3: Find the solution of $\sqrt{x-3}-\sqrt{x}=3$
Solution:

Here, there are two radicals, so squaring both the sides twice, we will get
$x-3 - 2\sqrt{x-3}\sqrt{x}+x = 9$ this implies
$2x-3 - 2\sqrt{x^{2}-3x} = 9$, or $x-6 = \sqrt{x^{2}-3x}$
Again squaring both sides,
$(x-6)^2 = (\sqrt{x^{2}-3x})^2$
$x^2 + 36 - 12x = x^2 - 3x$
36 - 9x = 0
x = 4
we will get x = 4 as the answer.

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