It’s given that $3x+\dfrac{2}{x}=7$, then find the value of $9{{x}^{2}}-\dfrac{4}{{{x}^{2}}}$.
A. 25 B. 35 C. 49 D. 30
Answer
612k+ views
Hint: The value of the equation we need to find is in the identity form of ${{a}^{2}}-{{b}^{2}}$. So, to find the value we need to find the value of its two factorised divisors. One is given in the form of $3x+\dfrac{2}{x}=7$. We need to find the other half. For that we use the given in the form of ${{\left( a+b \right)}^{2}}={{\left( a-b \right)}^{2}}+4ab$ to get the value. At the end we multiply to get the solution.
Complete step-by-step answer:
The given part is $3x+\dfrac{2}{x}=7$.
We square both side of the equation to get ${{\left( 3x+\dfrac{2}{x} \right)}^{2}}={{7}^{2}}=49$.
Now, we use the identity ${{\left( a+b \right)}^{2}}={{\left( a-b \right)}^{2}}+4ab$ on the previous equation.
We get ${{\left( 3x+\dfrac{2}{x} \right)}^{2}}={{\left( 3x-\dfrac{2}{x} \right)}^{2}}+4.\left( 3x \right).\left( \dfrac{2}{x} \right)={{\left( 3x-\dfrac{2}{x} \right)}^{2}}+24$.
So, the equation becomes ${{\left( 3x-\dfrac{2}{x} \right)}^{2}}+24=49$.
We solve it to get the solution of $3x-\dfrac{2}{x}$.
$\begin{align}
& {{\left( 3x-\dfrac{2}{x} \right)}^{2}}=49-24=25 \\
& \Rightarrow 3x-\dfrac{2}{x}=\pm 5 \\
\end{align}$
Now, we try to find the solution of $9{{x}^{2}}-\dfrac{4}{{{x}^{2}}}$. It’s an identity of the form ${{a}^{2}}-{{b}^{2}}$.
So, the factorisation form becomes ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$.
We put the identity to get $9{{x}^{2}}-\dfrac{4}{{{x}^{2}}}=\left( 3x+\dfrac{2}{x} \right)\left( 3x-\dfrac{2}{x} \right)$.
We put values to find the answer as $9{{x}^{2}}-\dfrac{4}{{{x}^{2}}}=\left( 3x+\dfrac{2}{x} \right)\left( 3x-\dfrac{2}{x} \right)=7\times \left( \pm 5 \right)=\pm 35$.
So, the value of $9{{x}^{2}}-\dfrac{4}{{{x}^{2}}}$ is $\pm 35$.
The only option matching with the answer is 35. So, (B) is the correct option.
Note: Although we have got 2 values, it can be that only 1 value is right. But we have no other condition to test it but if given then we have to check if both of them are valid or not. We can’t directly use the identity to find out the value of the $9{{x}^{2}}-\dfrac{4}{{{x}^{2}}}$. Whenever we are finding roots, we have to use both roots of different signs.
Complete step-by-step answer:
The given part is $3x+\dfrac{2}{x}=7$.
We square both side of the equation to get ${{\left( 3x+\dfrac{2}{x} \right)}^{2}}={{7}^{2}}=49$.
Now, we use the identity ${{\left( a+b \right)}^{2}}={{\left( a-b \right)}^{2}}+4ab$ on the previous equation.
We get ${{\left( 3x+\dfrac{2}{x} \right)}^{2}}={{\left( 3x-\dfrac{2}{x} \right)}^{2}}+4.\left( 3x \right).\left( \dfrac{2}{x} \right)={{\left( 3x-\dfrac{2}{x} \right)}^{2}}+24$.
So, the equation becomes ${{\left( 3x-\dfrac{2}{x} \right)}^{2}}+24=49$.
We solve it to get the solution of $3x-\dfrac{2}{x}$.
$\begin{align}
& {{\left( 3x-\dfrac{2}{x} \right)}^{2}}=49-24=25 \\
& \Rightarrow 3x-\dfrac{2}{x}=\pm 5 \\
\end{align}$
Now, we try to find the solution of $9{{x}^{2}}-\dfrac{4}{{{x}^{2}}}$. It’s an identity of the form ${{a}^{2}}-{{b}^{2}}$.
So, the factorisation form becomes ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$.
We put the identity to get $9{{x}^{2}}-\dfrac{4}{{{x}^{2}}}=\left( 3x+\dfrac{2}{x} \right)\left( 3x-\dfrac{2}{x} \right)$.
We put values to find the answer as $9{{x}^{2}}-\dfrac{4}{{{x}^{2}}}=\left( 3x+\dfrac{2}{x} \right)\left( 3x-\dfrac{2}{x} \right)=7\times \left( \pm 5 \right)=\pm 35$.
So, the value of $9{{x}^{2}}-\dfrac{4}{{{x}^{2}}}$ is $\pm 35$.
The only option matching with the answer is 35. So, (B) is the correct option.
Note: Although we have got 2 values, it can be that only 1 value is right. But we have no other condition to test it but if given then we have to check if both of them are valid or not. We can’t directly use the identity to find out the value of the $9{{x}^{2}}-\dfrac{4}{{{x}^{2}}}$. Whenever we are finding roots, we have to use both roots of different signs.
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