In solving a B-square equation in IR, we must substitute its variable into a 2nd degree equation. Notice now the procedure that should be used.

**Practical sequence**:

Replace x

^{4 }by y^{2}(or any other squared unknown) and x^{2 }by y.Solve the equation ay

^{2}+ by + c = 0.Determine the square root of each of the roots (y'e y ") of the equation ay

^{2}+ by + c = 0.

These two relationships tell us that each**positive root**of the equation ay^{2 }+ by + c = 0 gives rise to two symmetrical roots for the square: the**negative root**gives no real root to it.

Examples:

Determine the roots of the equation x

^{4}- 13x^{2}+ 36 = 0.**Solution**:

Replacing x^{4}by y^{2}and x^{2}by y we have:

y^{2 }- 13y + 36 = 0

By solving this equation we get:

y '= 4 and y "= 9

As x^{2}= y, we have:

So we have for truth set: V = {-3, -2, 2, 3}.

Determine the roots of the equation x

^{4}+ 4x^{2}- 60 = 0.**Solution:**Replacing x^{4}by y^{2}and x^{2}by y we have:

y^{2}+ 4y - 60 = 0

By solving this equation we get:

y '= 6 and y "= -10

As x^{2}= y, we have:So we have for the truth set:.

Determine the sum of the roots of the equation .

**Solution:**We use the following device:

Like this:

y^{2}- 3y = -2

y^{2}- 3y + 2 = 0

y '= 1 and y "= 2

Substituting y, we determine:

Therefore, the sum of the roots is given by:

## Resolution of equations of the form: ax2n + bxno + c = 0

This kind of equation can be solved in the same way as the square one. For that, we replace x^{no} by y getting:

ay^{2} + by + c = 0, which is an equation of the 2nd degree.

Example:

Solve equation x

^{6}+ 117x^{3}- 1.000 = 0.**Solution**:

Doing x^{3}= y, we have:

y^{2}+ 117y - 1,000 = 0Solving the equation, we get:

y '= 8 and y "= - 125

So:

Thus, V = {-5,2}.