Question

How do you factor the equation \[3{{x}^{4}}-2{{x}^{2}}-1=0\]?

Answer 1

Hint: Now to solve this equation we will first substitute ${{x}^{2}}=t$ . Now we have a quadratic in t. we will solve the quadratic by completing the square method. Hence we will first divide the whole equation by a and then add and subtract the term ${{\left( \dfrac{b}{2a} \right)}^{2}}$ . Now we will simplify the equation by using ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ and then take square root to find the value of t. Now re-substituting the value of t we will find the value of x.

Complete step by step solution:
Now we are given with an equation of degree 4 in x.
To solve the equation let us substitute ${{x}^{2}}=t$ Hence squaring on both sides we get, ${{x}^{4}}={{t}^{2}}$
Hence we get the given expression as $3{{t}^{2}}-2t-1=0$ .
Now we have a quadratic equation in t.
Now we will try to find the roots of the expression.
Let us use completing the square method to solve the equation.
Now to use this method we will first make the coefficient of ${{t}^{2}}$ as 1.
Hence dividing the whole equation by 3 we get, ${{t}^{2}}-\dfrac{2}{3}t-\dfrac{1}{3}=0$
Now we will add and subtract ${{\left( \dfrac{-2}{3\left( 2 \right)} \right)}^{2}}=\dfrac{4}{36}$ to the equation.
Hence we get, ${{t}^{2}}-\dfrac{2}{3}t+\dfrac{4}{36}-\dfrac{4}{36}-\dfrac{1}{3}=0$ .
Now we know that ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ . Hence using this we get,
$\Rightarrow {{\left( t-\dfrac{2}{6} \right)}^{2}}-\dfrac{4}{36}-\dfrac{1}{3}=0$
Now taking LCM in the equation we get,
$\begin{align}
  & \Rightarrow {{\left( t-\dfrac{2}{6} \right)}^{2}}+\dfrac{-4-12}{36}=0 \\
 & \Rightarrow {{\left( t-\dfrac{2}{6} \right)}^{2}}=\dfrac{16}{36} \\
\end{align}$
Now taking square root in the equation we get,
$\Rightarrow \left( t-\dfrac{2}{6} \right)=\pm \dfrac{4}{6}$
Now taking $\dfrac{2}{6}$ to RHS we get,
$\Rightarrow t=\dfrac{2}{6}\pm \dfrac{4}{6}$
Now hence we have $t=1$ or $t=\dfrac{-1}{3}$
Now re-substituting the value of x we get
$\Rightarrow {{x}^{2}}=1$ or ${{x}^{2}}=\dfrac{-1}{3}$
Hence taking the square root we have x = 1, x = -1, x = $\dfrac{-i}{3}$ and $x=\dfrac{i}{3}$ .
Hence the factors of the equation are $\left( x-1 \right)\left( x+1 \right)\left( x+\dfrac{i}{3} \right)\left( x-\dfrac{i}{3} \right)$

Note: Now note that for a quadratic equation of the form $a{{x}^{2}}+bx+c=0$ we can find the roots by using the direct formula. Now the roots of these equations are given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . Hence we can use this formula to find the roots of the equation.