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Solve the following quadratic equation by factorization:${{a}^{2}}{{b}^{2}}{{x}^{2}}+{{b}^{2}}x-{{a}^{2}}x-1=0$ Now, state if the roots of the above equation are $-\dfrac{1}{{{a}^{2}}},\dfrac{1}{{{b}^{2}}}$ is:(a) True(b) False

Last updated date: 20th Jun 2024
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Hint: In the above problem, we are asked to solve the above quadratic equation ${{a}^{2}}{{b}^{2}}{{x}^{2}}+{{b}^{2}}x-{{a}^{2}}x-1=0$ by factorization. For that we will take some common term from the first two terms of the above equation and some common factor from the last two terms and then simplify the expression to completely factorize the above equation. After that, we will equate each factor to 0 to get the value of x.

Complete step-by-step solution:
The quadratic equation given above of which we have to find solutions are as follows:
${{a}^{2}}{{b}^{2}}{{x}^{2}}+{{b}^{2}}x-{{a}^{2}}x-1=0$
As you can see that in the first two terms ${{b}^{2}}x$ is a common expression so taking ${{b}^{2}}x$ as common from the first two expressions we get,
$\Rightarrow {{b}^{2}}x\left( {{a}^{2}}x+1 \right)-{{a}^{2}}x-1=0$
Now, taking -1 as common from the last two terms of the above equation we get,
$\Rightarrow {{b}^{2}}x\left( {{a}^{2}}x+1 \right)-1\left( {{a}^{2}}x+1 \right)=0$
As you can see that ${{a}^{2}}x+1$ is common in the above equation so we can take ${{a}^{2}}x+1$ as common then we get,
$\Rightarrow \left( {{a}^{2}}x+1 \right)\left( {{b}^{2}}x-1 \right)=0$
Now, equating each of the brackets to 0 to get the solutions of x we get,
\begin{align} & \Rightarrow {{a}^{2}}x+1=0 \\ & \Rightarrow x=-\dfrac{1}{{{a}^{2}}} \\ \end{align}
\begin{align} & \Rightarrow {{b}^{2}}x-1=0 \\ & \Rightarrow x=\dfrac{1}{{{b}^{2}}} \\ \end{align}
Hence, we have found the solutions of the above quadratic equation as:
$\Rightarrow x=-\dfrac{1}{{{a}^{2}}},\dfrac{1}{{{b}^{2}}}$
As we are getting the same roots as given in the above problem so the correct answer is True.

Note: We can verify the solutions in x which we are getting above by substituting the above values of x and then see whether those values of x are satisfying the above equation or not.
We are checking the root $x=-\dfrac{1}{{{a}^{2}}}$ by substituting this value of x in the above quadratic equation we get,
\begin{align} & \Rightarrow {{a}^{2}}{{b}^{2}}{{\left( -\dfrac{1}{{{a}^{2}}} \right)}^{2}}+{{b}^{2}}\left( -\dfrac{1}{{{a}^{2}}} \right)-{{a}^{2}}\left( -\dfrac{1}{{{a}^{2}}} \right)-1=0 \\ & \Rightarrow {{a}^{2}}{{b}^{2}}\left( \dfrac{1}{{{a}^{4}}} \right)-\dfrac{{{b}^{2}}}{{{a}^{2}}}+1-1=0 \\ \end{align}
In the above equation, ${{a}^{2}}$ will be cancelled out in the numerator and denominator of the first term written in the L.H.S then we get,
\begin{align} & \Rightarrow {{b}^{2}}\left( \dfrac{1}{{{a}^{2}}} \right)-\dfrac{{{b}^{2}}}{{{a}^{2}}}+1-1=0 \\ & \Rightarrow \dfrac{{{b}^{2}}}{{{a}^{2}}}-\dfrac{{{b}^{2}}}{{{a}^{2}}}+0=0 \\ & \Rightarrow 0+0=0 \\ & \Rightarrow 0=0 \\ \end{align}
As you can see that L.H.S = R.H.S so the value of $x=-\dfrac{1}{{{a}^{2}}}$ is satisfying the above equation.
Similarly, you can check the other solution of x too.