# Equations – Solving a polynomial equation – Exercise 5581

Exercise

Solve the equation:

$$5x^4+3x^2-8=0$$

$$x=\pm 1$$

Solution

$$5x^4+3x^2-8=0$$

In order to get quadratic equation, we define a new variable:

$$y=x^2$$

We set the new variable:

$$5y^2+3y-8=0$$

It’s a quadratic equation with the coefficients:

$$a=5, b=3, c=-8$$

We solve it with the quadratic formula. Putting the coefficients in the formula gives us

$$y_{1,2}=\frac{-3\pm \sqrt{3^2-4\cdot 5\cdot (-8)}}{2\cdot 5}=$$

$$=\frac{-3\pm \sqrt{169}}{10}=$$

$$=\frac{-3\pm 13}{10}$$

Hence, we get the solutions:

$$y_1=\frac{-3+ 13}{10}=\frac{10}{10}=1$$

$$y_2=\frac{-3- 13}{10}=\frac{-16}{10}=-1.6$$

We go back to the original variable. From the first solution we get

$$1=x^2$$

$$x=\pm 1$$

From the second solution we get

$$-1.6=x^2$$

This equation has no real solution, because the left side is negative, while the right side is positive for every x.

Hence, the only solution of the equation is

$$x=\pm 1$$

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