Question

4 Solve the biquadratic equations:(d) x^2 (4x^2 - 13) = -3

Answer 1

Answer::

`x=-0.5,0.5,-√(3),√(3)`

Explanation:

Given the biquadratic equation:

`x^2(4x^2-13)=-3`

First, add 3 to both sides of the equation:

`\begin{gathered} x^2(4x^2-13)+3=-3+3 \\ x^2(4x^2-13)+3=0 \end{gathered}`

Next, open the bracket:

`4x^4-13x^2+3=0`

Factorize the resulting expression:

`\begin{gathered} 4x^4-12x^2-x^2+3=0 \\ 4x^2(x^2-3)-1(x^2-3)=0 \\ (4x^2-1)(x^2-3)=0 \end{gathered}`

Then, solve the equation for x:

`\begin{gathered} 4x^2-1=0,x^2-3=0 \\ 4x^2=1,x^2=3 \\ x^2=(1)/(4),x^2=3 \\ x=\pm\sqrt{(1)/(4)},x=\pm√(3) \\ x=-0.5,0.5,-√(3),√(3) \end{gathered}`

The solution to the biquadratic equations are:

`x=-0.5,0.5,-√(3),√(3)`