Question

How do you solve \[\dfrac{{3x}}{{x + 1}} = \dfrac{{12}}{{{x^2} - 1}} + 2\]?

Answer 1

Hint:To solve this question first we take all the variable parts on one side and then we multiply by \[{x^2} - 1\] on both sides. Then we cancel the fraction from the numerator and denominator which is cancelable. Then we simplify that equation and make a quadratic equation and then find the roots of the quadratic equation. That is the value of x satisfying the given equation.


Complete step by step answer:
We have given, \[\dfrac{{3x}}{{x + 1}} = \dfrac{{12}}{{{x^2} - 1}} + 2\]
To find,
The value of the variable is \[x\].
\[\dfrac{{3x}}{{x + 1}} = \dfrac{{12}}{{{x^2} - 1}} + 2\] ……(i)
On taking all the variables in one side.
\[\dfrac{{3x}}{{x + 1}} - \dfrac{{12}}{{{x^2} - 1}} = 2\]
Multiplying by \[{x^2} - 1\] on both sides.
\[\dfrac{{3x\left( {{x^2} - 1} \right)}}{{x + 1}} - \dfrac{{12\left( {{x^2} - 1} \right)}}{{{x^2} - 1}} = 2\left( {{x^2} - 1} \right)\]
Now cancelling the common factor.
\[\dfrac{{3x\left( {{x^2} - 1} \right)}}{{x + 1}} - 12 = 2\left( {{x^2} - 1} \right)\]
Factoring the numerator part of the first term.
\[\dfrac{{3x\left( {x - 1} \right)\left( {x + 1} \right)}}{{x + 1}} - 12 = 2\left( {{x^2} - 1} \right)\]
Again canceling the common factor.
\[3x\left( {x - 1} \right) - 12 = 2\left( {{x^2} - 1} \right)\]
Now multiplying the terms.
\[3{x^2} - 3x - 12 = 2{x^2} - 2\]
Taking all the parts on one side of the equal sign.
\[3{x^2} - 2{x^2} - 3x - 12 + 2 = 0\]
On further calculating
\[{x^2} - 3x - 10 = 0\]
Now for finding the value x we have to find the solutions of the obtained quadratic equation.
We solve this equation by the factorization method.
So, we will split the middle term.
\[{x^2} - 5x + 2x - 10 = 0\]
On taking x common from the first two terms and 2 from next two terms.
\[x\left( {x - 5} \right) + 2\left( {x - 5} \right) = 0\]
Taking \[x - 5\] common from both the terms.
\[\left( {x + 2} \right)\left( {x - 5} \right) = 0\]
On both the terms equal to 0 we get the value of \[x\].
\[x + 2 = 0\] and \[x - 5 = 0\]
Now the values of \[x\] are \[ - 2\] and 5
On solving equation \[\dfrac{{3x}}{{x + 1}} = \dfrac{{12}}{{{x^2} - 1}} + 2\] we get the values of \[x\] are \[ - 2\] and 5.

Note:This question is not too lengthy and not hard. But students commit mistakes in solving the equation. Students often make mistakes in finding the roots of the quadratic equation. Roots of the quadratic equation and found by many methods like by perfect square method and one simplest method is by putting all the values in \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] formula.