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# Solve for $x$: $12ab{x^2} - \left( {9{a^2} - 8{b^2}} \right)x - 6ab = 0$.

Last updated date: 20th Jun 2024
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Hint: In this question, we are asked to solve the given quadratic equation which means that we have to find the roots of this given quadratic equation and that can be done by using the quadratic formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ or we will factorize the equation by splitting the middle term. Split the term such that the product of these terms is the product of first and constant terms. Then, take the terms common and represent the equation as a product of two factors.

Complete step-by-step solution:
As mentioned in the question, we have to find the value of $x$ which is the roots of the quadratic equation.
Now the quadratic equation is as follows:
$\Rightarrow 12ab{x^2} - \left( {9{a^2} - 8{b^2}} \right)x - 6ab = 0$
Now, on using the quadratic formula, we get the following result
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Here, $a = 12ab$, $b = - \left( {9{a^2} - 8{b^2}} \right)$ and $c = - 6ab$
Substitute the values in the equation,
$\Rightarrow x = \dfrac{{ - \left\{ { - \left( {9{a^2} - 8{b^2}} \right)} \right\} \pm \sqrt {{{\left\{ { - \left( {9{a^2} - 8{b^2}} \right)} \right\}}^2} - 4 \times 12ab \times - 6ab} }}{{2 \times 12ab}}$
We know that,
${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$
Use the above formula to expand the term and simplify,
Simplify the terms,
$\Rightarrow x = \dfrac{{\left( {9{a^2} - 8{b^2}} \right) \pm \sqrt {81{a^4} + 64{b^4} - 144{a^2}{b^2} + 288{a^2}{b^2}} }}{{24ab}}$
Subtract the like terms in the square root,
$\Rightarrow x = \dfrac{{\left( {9{a^2} - 8{b^2}} \right) \pm \sqrt {81{a^4} + 64{b^4} + 144{a^2}{b^2}} }}{{24ab}}$
We know that,
${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$
Use the formula to make the square inside the square root,
$\Rightarrow x = \dfrac{{\left( {9{a^2} - 8{b^2}} \right) \pm \sqrt {{{\left( {9{a^2} + 8{b^2}} \right)}^2}} }}{{24ab}}$
Cancel out the square with square root,
$\Rightarrow x = \dfrac{{\left( {9{a^2} - 8{b^2}} \right) \pm \left( {9{a^2} + 8{b^2}} \right)}}{{24ab}}$
Simplify the terms,
$\Rightarrow x = \dfrac{{\left( {9{a^2} - 8{b^2}} \right) + \left( {9{a^2} + 8{b^2}} \right)}}{{24ab}},\dfrac{{\left( {9{a^2} - 8{b^2}} \right) - \left( {9{a^2} + 8{b^2}} \right)}}{{24ab}}$
$\Rightarrow x = \dfrac{{18{a^2}}}{{24ab}},\dfrac{{ - 16{b^2}}}{{24ab}}$
Cancel out the common terms,
$\therefore x = \dfrac{{3a}}{{4b}}, - \dfrac{{2b}}{{3a}}$

Hence, the value of $x$ is $\dfrac{{3a}}{{4b}}, - \dfrac{{2b}}{{3a}}$.

Note: This question can be done in another way also.
We have to factorize the quadratic equation
$\Rightarrow 12ab{x^2} - \left( {9{a^2} - 8{b^2}} \right)x - 6ab = 0$
Here the middle term is already split such that the sum of the terms is the middle term and the product of the terms such as the product of the first term and constant term of the original equation.
Then, the equation can be written as
$\Rightarrow 12ab{x^2} - 9{a^2}x + 8{b^2}x - 6ab = 0$
Take $3ax$ common from the first two terms and $2b$ common from the last two terms.
$\Rightarrow 3ax\left( {4bx - 3a} \right) + 2b\left( {4bx - 3a} \right) = 0$
Now, take $\left( {4bx - 3a} \right)$ common from both terms,
$\Rightarrow \left( {3ax + 2b} \right)\left( {4bx - 3a} \right) = 0$
Equate each term with 0,
$\Rightarrow 3ax + 2b = 0$ and $4bx - 3a = 0$
Move constant part on the other side,
$\Rightarrow 3ax = - 2b$ and $4bx = 3a$
Simplify the terms,
$\therefore x = \dfrac{{3a}}{{4b}}, - \dfrac{{2b}}{{3a}}$
Hence, the value of $x$ is $\dfrac{{3a}}{{4b}}, - \dfrac{{2b}}{{3a}}$.