Question

Solve the given polynomial equation for x $x:9{{x}^{4}}-27{{x}^{2}}-36=0$
(a) $9\left( {{x}^{2}}-4 \right)\left( {{x}^{2}}+1 \right)$
(b) $\left( {{x}^{2}}-4 \right)\left( 9{{x}^{2}}+1 \right)$
(c) $9\left( {{x}^{2}}+4 \right)\left( {{x}^{2}}-1 \right)$
(d) $9\left( {{x}^{2}}-4 \right)\left( {{x}^{2}}-1 \right)$

Answer 1

- Hint: Suppose the term ${{x}^{2}}$ as t in the given equation and get the quadratic equation in terms of ‘t’. Now, factorize this equation by mid-term splitting i.e. split the mid-term (term with coefficient ‘t’) into two terms, such that multiplication of both terms equals to multiplication of first and last term. Now, put $t={{x}^{2}}$ to get the required answer.

Complete step-by-step solution -

Given expression in the problem is
$9{{x}^{4}}-27{{x}^{2}}-36=0.................\left( i \right)$
Now, as we can observe that the given expression is a biquadratic polynomial i.e. degree of the polynomial is 4.
So, we can rewrite the equation (i) by replacing ${{x}^{4}}$ by ${{\left( {{x}^{2}} \right)}^{2}}$ as
$9{{\left( {{x}^{2}} \right)}^{2}}-27{{x}^{2}}-36=0...................\left( ii \right)$
Now, suppose the term ${{x}^{2}}$ as t i.e.
$t={{x}^{2}}................\left( iii \right)$
Hence, we can rewrite the equation (ii) in terms of ‘t’ using equation (iii) as
$9{{t}^{2}}-27t-36=0$
Now, we can divide the above equation by 9 as all the terms are divisible by 9. So, we get the above quadratic by taking ‘9’ as common.
We get
$9\left( {{t}^{2}}-3t-4 \right)=0...............\left( iv \right)$
Now, we have to factorize the above relation using mid-term splitting i.e. we need to split -3t into two terms such that multiplication of them is equal to the multiplication of first and last term of the equation. So, we can write -3t as the sum of ‘-4t’ and ‘t’. Hence, we can rewrite the equation (iv) as
$9\left( {{t}^{2}}-4t+t-4 \right)=0$
Taking ‘t’ as common from the first two terms and ‘1’ from the last two terms. So, we get
$\begin{aligned} & 9\left[ t\left( t-4 \right)+1\left( t-4 \right) \right]=0 \\\ & 9\left( t-4 \right)\left( t+1 \right)=0.................\left( v \right) \\\ \end{aligned}$
Now, we can get the above equation in terms of $'x'$ using equation (iii) i.e. $t={{x}^{2}},$ as
$9\left( {{x}^{2}}-4 \right)\left( {{x}^{2}}+1 \right)=0$
Hence, we can factorize the equation $9{{x}^{4}}-27{{x}^{2}}-36=0$ as $9\left( {{x}^{2}}-4 \right)\left( {{x}^{2}}+1 \right)=0$
Hence, option (a) in the correct answer of the problem.

Note: We can factorize the calculated answer $9\left( {{x}^{2}}-4 \right)\left( {{x}^{2}}+1 \right)$ further as well with the help of algebraic identity $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right)$
So, we get
$\begin{aligned} & =9\left( {{\left( x \right)}^{2}}-{{\left( 2 \right)}^{2}} \right)\left( {{x}^{2}}+1 \right) \\\ & =9\left( x-2 \right)\left( x+2 \right)\left( {{x}^{2}}+1 \right) \\\ \end{aligned}$
One may directly factorize the given equation with supposing ${{x}^{2}}$ as t. It is done for the better understanding of the problem. It may be complex for some students to factorize the relation in ${{x}^{2}}$ directly.