Question

Find all the roots of the following equation y=x^4-7x^2+12

Answer 1

The given expression is :

`y=x^4-7x^2+12`

To find the roots :

`\begin{gathered} \text{ Let u=x}^2\text{ then } \\ u^2=x^4 \end{gathered}`

So, the equation will be :

`x^4-7x^2+12=u^2-7u+12`

Simplify the u by factorization method :

`\begin{gathered} u^2-7u+12=u^2-3u-4u+12 \\ u^2-7u+12=u(u-3)-4(u-3) \\ u^2-7u+12=(u-4)(u-3)_{} \end{gathered}`

as : u = x² , substitute these value back :

`\begin{gathered} u^2-7u+12=(u-4)(u-3) \\ u^2-7u+12=(x^2-4)(x^2-3) \end{gathered}`

Apply the quadratic expression : (a² - b²) = (a - b)(a + b)

`\begin{gathered} u^2-7u+12=\mleft(x^2-4\mright)\mleft(x^2-3\mright) \\ x^4^{}-7x^2+12=(x^2-2^2_{})(x^2-\sqrt[]{3}^2) \\ x^4-7x^2+12=(x-2)(x+2)(x-\sqrt[]{3})(x+\sqrt[]{3}) \end{gathered}`

`\text{ So, the factors are : }=(x-2)(x+2)(x-\sqrt[]{3})(x+\sqrt[]{3})`

For the roots, equate each factor with zero:

`\begin{gathered} (x-2)=0\text{ }\Rightarrow x=2 \\ (x+2)=0\Rightarrow x=-2 \\ (x-\sqrt[]{3})=0\Rightarrow x=\sqrt[]{3} \\ (x+\sqrt[]{3})=0\Rightarrow x=-\sqrt[]{3} \end{gathered}`

So, the roots are :

`x=2,\text{ -2, }\sqrt[]{3},-\sqrt[]{3}`

Answer :

`x=2,\text{ -2, }\sqrt[]{3},-\sqrt[]{3}`