solving radical equations

Radical equations

A radical equation is an equation in which the variable appears under a radical sign or has a fractional exponent, since fractional exponents are an alternate way of expressing radicals or roots.

The radical sign in a radical equation always indicates
the principal or positive root of the expression under it.

Because of this fact, we must always check our solutions for a radical equation because sometimes it is the negative root of the solution which makes the equation a true statement, and we must reject that solution. Let's do an example to illustrate this situation and then we'll list the steps for solving radical equations.

solve for x

1) Isolate the radical

2) Square both sides

3) Solve

4) Check the solution

This last statement would be true if we used – 2 as the root of 4.

But we are limited to using the principal or positive square root of 4, so x = 3 is not a root of this equation. If the sign before the radical in the original equation had been a negative, then 3 would be a solution for this equation.

This problem arises because of step 2. When we square both sides of the equation we are performing an operation which could make an untrue statement true.

For example, the statement 1 = +1 is untrue. However, if we square both sides of this untruth, it becomes a true statement because (1) 2 does equal (+1) 2. Therefore we must always check our solutions for radical equations.

Steps for solving radical equations

1) Isolate a radical on one side of the equation.

2) Square both sides of the equation.

3) Collect like terms.

4) If there are no longer any radicals in the equation, solve it.

5) If there are still radicals in the equation, go back to step 1.

Note: When we say square both sides, we mean exactly that. Each side of the equation is taken as a unit. The most common error made is to square the individual terms rather than the complete side of the equation. To illustrate that this is not the way to proceed, let's do a simple numerical example. Say we have the true statement

If we square the individual terms, we will get 9 + 25 = 64 which of course is ridiculous since we know that 9 + 25 = 34.

However, squaring both sides of the equation using our method of squaring a binomial yields the true statement that 9 + 2(3)(5) + 25 = 64. As you can see in the previous example, when we square the left side of the equation, we get a trinomial because we squared a binomial.

When, there's more than one radical in the original equation, we perform the steps listed above until there are no radicals left. Let's do some examples.


practice examples solutions



solve these radical equations:


substitute x = 4: ü

substitute x = ¾

so x = ¾ is not a solution.

Isolate the radical term.

Square both sides of the equation.

Collect like terms, relate to zero.


Solve for x.

Check the solutions. As shown, x = 4 works but x = ¾ doesn't make the original equation a true statement. The only solution for this radical equation is x = 4.


substitute a = 5; ü

b) Isolate the radicals.

Square both sides of the equation.

Solve for a and check the solution in the

original equation. As shown, it works.


substitute y = 4; ü

substitute y = 38; ü

c) Isolate one of the radicals on the left


Square both sides of the equation.

Collect terms. Divide through by 2.

Square both sides and solve.

Check the solution.


substitute x = 2; ü

d) Isolate a radical on the left side.

Square both sides of the equation.

Collect like terms.

Square both sides of the equation again.

Collect like terms.

Factor and solve.

Check the solutions in the original equation.

practice examples solutions


solve these radical equations:







practice examples solutions



25 = y + 10

15 = y

check: ü


m2 28m + 196 = 16m + 112

m2 44m + 84 = 0

(m 42)(m 2) = 0

m = 42 or m = 2

check: m = 42,

check: m = 2, it works. So solution is m = 2.


4x + 5 = 9x2 + 24x + 16

0 = 9x2 + 20x + 11

0 = (9x + 11)(x + 1)

x = 11/9 or x = 1

both solutions work.


16 = y + 7 so y = 9 and it works.


16x = 144 24x + x2

x2 40x + 144 = 0

(x 4)(x 36) = 0 so x = 4 or x = 36

Check: x = 4 works but x = 36 does not.


x2 6x + 9 = 4(x 3)

x2 10x + 21 = 0

(x 7)(x 3) = 0, so x = 7 or x = 3

both solutions work.

For word problems involving radicals, see the square root function in the Functions MathRoom.

practice examples solutions

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