# Equations – Solving a polynomial equation – Exercise 5584

Exercise

Solve the equation

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

$$x=\pm 1, x=\pm \sqrt{2}$$

Solution

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

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

$$y=x^2$$

We set the new variable:

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

It’s a quadratic equation with the coeffients:

$$a=1, b=-3, c=2$$

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

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

$$=\frac{3\pm \sqrt{1}}{2}=$$

$$=\frac{3\pm 1}{2}$$

Hence, we get the solutions:

$$y_1=\frac{3+ 1}{2}=\frac{4}{2}=2$$

$$y_2=\frac{3- 1}{2}=\frac{2}{2}=1$$

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

$$2=x^2$$

$$x=\pm \sqrt{2}$$

From the second solution we get

$$1=x^2$$

$$x=\pm 1$$

Finally, the solutions of the equation are

$$x=\pm 1,\pm\sqrt{2}$$

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