Do you need to solve biquadratic equations? Use our online calculator and get the four solutions of the equation immediately.

Recall that a biquadratic equation are fourth degree equations (not to be confused with second degree equations) without any terms of odd degree. That is, they look like this:

ax4 + bx2 + c = 0

To use our biquadratic equations calculator calculator just type the values of a, b and c and press the calculate button to get the result.

## How to solve biquadratic equations

We have already seen that a two-square equation is of the form:

ax4 + bx2 + c = 0

To solve it, what we are going to do is transforming it into a second degree equation that we do know how to solve. To do this, we will make a change of variable in which:

x2 = t

Therefore, if we change the variable in the two-square equation, we will have a second degree equation of the form:

at2 + bt + c = 0

The second degree equations we already know how to solve it (click on the link we have just left if you don't know how to do it) so we will get two solutions from it:

t = t1

t = t2

But since we made a change of variable in which x2 = t, we will have that to draw out the four solutions of the biquadratic equation by solving the following square roots:

x1 = +√ t1
x2 = -√ t1
x3 = +√ t2
x4 = -√ t2

Transforming a two-square root into a second degree equation makes its calculation much easier.

## Exercises of two-square equations solved

In order to put into practice the theory we have just seen with biquadratic equations exercisesLet's solve the following equation:

x4 - 16x2 - 225 = 0

We make the change of variable x2 = t and transform it into the following second degree equation:

t2 – 16t – 225 = 0

To solve it, we will apply the general formula which allows us to obtain the solutions of a second degree equation: Making substitutions in the above formula and solving, we are left with the following: Now that we know the solutions, we undo the variable change we made at the beginning of the article:

x1 = +√ t1 = +√ 25 = 5
x2 = -√ t1 = -√ 25 = -5
x3 = +√ t2 = +√ -9
x4 = -√ t2 = -√ -9

We do not solve the third and fourth solutions since there are no real numbers of the square root of -9 (we can only do the square root of positive numbers). Leaving this indicated in the exercise is enough.

In case the previous exercise of two-square equations has not been very clear, let's see another one in which we are asked to solve:

x4 - 13x2 + 36 = 0

We change the variable x2 = t to transform it into a second degree equation:

t2 – 13t + 36 = 0

And we solve it: Finally, we undo the change of variable and extract the four solutions of the initial two-square equation:

x1 = +√ t1 = +√ 9 = 3
x2 = -√ t1 = -√ 9 = -3
x3 = +√ t2 = +√ 4 = 2
x4 = -√ t2 = -√ 4 = -2

Easy, isn't it? If you have any doubts or want to check the solutions, you can always use our biquadratic equation calculator or leave us a comment and we will help you solve the exercise.

## How many solutions does a two-square equation have? Although a two-square equation normally has four solutionsIn some cases, they may have three, two, a single solution or even none at all.

For example:

• If t1 = 0 or t2 = 0, we will only have a single solution when undoing the change of variable and it will be x = 0
• If t1 < 0 we will not have any solution as we have seen in the first example since there are no roots of a negative number. It would be the same if t2 < 0.

If you use our biquadratic equation calculator and you see that in some of the fields of the solutions no data is displayed, it is because some of the above conditions have been met.

## Solving incomplete two-square equations When it comes to solving a incomplete two-square equationThe procedure to be used is the same as in the first point of the theory.

In other words, we change the variable and as a result we will obtain a incomplete quadratic equation that will have to be solved following the procedure that we indicate in the page of the link that we have just left you.