Question

Solve the following equation:
\[6{{x}^{4}}-25{{x}^{3}}+12{{x}^{2}}+25x+6=0\]

Answer 1

Hint: At first, rearrange the given equation and write it as, \[6{{x}^{4}}-12{{x}^{2}}+6-25{{x}^{3}}+25x+24{{x}^{2}}=0.\] After that, write it as \[6{{\left( {{x}^{2}}-1 \right)}^{2}}-25x\left( {{x}^{2}}-1 \right)+24{{x}^{2}}=0.\] Then transform it into two quadratic equation factors and factorize them separately to find the values of x.

Complete step-by-step answer:
In this question, we are given an equation \[6{{x}^{4}}-25{{x}^{3}}+12{{x}^{2}}+25x+6=0\] and we have to find for the values of x so that it satisfies the given expression. The equation given in the question is
\[6{{x}^{4}}-25{{x}^{3}}+12{{x}^{2}}+25x+6=0\]
It can be further written as,
\[6{{x}^{4}}-12{{x}^{2}}+6-25{{x}^{3}}+25x+24{{x}^{2}}=0\]
\[\Rightarrow 6\left( {{x}^{4}}-2{{x}^{2}}+1 \right)-25x\left( {{x}^{2}}-1 \right)+24{{x}^{2}}=0\]
We know that,
\[{{\left( {{x}^{2}}-1 \right)}^{2}}={{x}^{4}}-2{{x}^{2}}+1\]
So, we can write the equation as,
\[6{{\left( {{x}^{2}}-1 \right)}^{2}}-25x\left( {{x}^{2}}-1 \right)+24{{x}^{2}}=0\]
The above equation can be further written as,
\[6{{\left( {{x}^{2}}-1 \right)}^{2}}-9x\left( {{x}^{2}}-1 \right)-16x\left( {{x}^{2}}-1 \right)+24{{x}^{2}}=0\]
On further simplification, we get,
\[3\left( {{x}^{2}}-1 \right)\left[ 2\left( {{x}^{2}}-1 \right)-3x \right]-8x\left[ 2\left( {{x}^{2}}-1 \right)-3x \right]=0\]
Hence, it can be factorized into,
\[\left[ 2\left( {{x}^{2}}-1 \right)-3x \right]\left[ 3\left( {{x}^{2}}-1 \right)-8x \right]=0\]
So, we can write it as,
\[\left( 2{{x}^{2}}-3x-2 \right)\left( 3{{x}^{2}}-8x-3 \right)=0\]
Now, we can write it as,
\[\left( 2{{x}^{2}}-4x+x-2 \right)\left( 3{{x}^{2}}-9x+x-3 \right)=0\]
This can be further written as,
\[\left[ 2x\left( x-2 \right)+1\left( x-2 \right) \right]\left[ 3x\left( x-3 \right)+1\left( x-3 \right) \right]=0\]
So, it can be factored into,
\[\left( 2x+1 \right)\left( x-2 \right)\left( 3x+1 \right)\left( x-3 \right)=0\]
Hence, the value of x for which the equation satisfies is
\[x=\dfrac{-1}{2},2,\dfrac{-1}{3},3\]
The value of x are \[\dfrac{-1}{2},2,\dfrac{-1}{3},3.\]


Note: One can also do the same question by another method. At first, divide the whole question by \[{{x}^{2}}\] and the equation formed will be \[6{{x}^{2}}-25x+12+\dfrac{25}{x}+\dfrac{6}{{{x}^{2}}}=0.\] Then take \[x-\dfrac{1}{x}\] as a variable t and so we can write the equation as \[6\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}}-2 \right)-25\left( x-\dfrac{1}{x} \right)+25=0\] as \[{{t}^{2}}=\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}}-2 \right).\] Then factorize the quadratic equation to find t and hence we will get the value of x.