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Find the real or imaginary solutions. x^4 + 3x^2 - 18 = 0

User Meeka
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1 Answer

5 votes

Answer:


x_1=\mathbf{i}√(6),\ x_2=-\mathbf{i}√(6),\ x_3=√(3),\ x_4=-√(3)

Explanation:

Biquadratic Equations

Solve:


x^4 + 3x^2 - 18 = 0

The biquadratic equations are equations of degree 4 without the terms of degree 1 and 3.

Solving such equations requires to express the equation as a second-degree equation with
x^2 as the variable.

Rewriting the equation:


(x^2)^2 + 3(x^2) - 18 = 0

The quadratic equation can be factored as:


(x^2+6)(x^2-3)=0

It leads to two equations:


x^2+6=0


x^2-3=0

The first equation has imaginary roots. Solving for x:


x^2=-6


x=\pm√(-6)


x_1=\mathbf{i}√(6)


x_2=-\mathbf{i}√(6)

Where


\mathbf{i}=√(-1)

The second equation has two real roots:


x^2-3=0


x^2=3


x=\pm√(3)


x_3=√(3)


x_4=-√(3)

The roots are:


\mathbf{x_1=\mathbf{i}√(6),\ x_2=-\mathbf{i}√(6),\ x_3=√(3),\ x_4=-√(3)}

User Collin James
by
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