Question

If we have an algebraic equation as ${{x}^{2}}+\dfrac{1}{{{x}^{2}}}=27$ , then find the value of $x-\dfrac{1}{x}$ .

Answer 1

Hint: Use the formula ${{\left( a-b \right)}^{2}}+2ab={{a}^{2}}+{{b}^{2}}$ , for simplification of the left-hand side of the equation ${{x}^{2}}+\dfrac{1}{{{x}^{2}}}=27$ followed by the rooting of both the side of the equation to reach the required result.

Complete step-by-step solution -
It is given in the question that:
 $\therefore {{x}^{2}}+\dfrac{1}{{{x}^{2}}}=27$
We know that ${{\left( a-b \right)}^{2}}+2ab={{a}^{2}}+{{b}^{2}}$ . So, if we use this to simplify the right-hand side of our equation, we get
$\Rightarrow {{\left( x-\dfrac{1}{x} \right)}^{2}}+2x\times \dfrac{1}{x}=27$
$\Rightarrow {{\left( x-\dfrac{1}{x} \right)}^{2}}+2=27$
If we take 2 to the other side of the equation, we get
${{\left( x-\dfrac{1}{x} \right)}^{2}}=27-2$
$\Rightarrow {{\left( x-\dfrac{1}{x} \right)}^{2}}=25$
Now we know that ${{a}^{2}}=b$ implies $a=\pm \sqrt{b}$ . So, our equation becomes:
$x-\dfrac{1}{x}=\pm \sqrt{25}$
We know that 25 is a perfect square and its square root comes out to be equal to 5. Therefore, we can say that
$x-\dfrac{1}{x}=\pm \sqrt{25}$
$\Rightarrow x-\dfrac{1}{x}=\pm 5$
So, there are two possible values of $x-\dfrac{1}{x}$ , which are 5 and -5. Hence, the answer to the above question is 5 and -5.

Note: We could have also solved the above question by changing the equation ${{x}^{2}}+\dfrac{1}{{{x}^{2}}}=27$ into a biquadratic polynomial by multiplying ${{x}^{2}}$ to both sides of the equation in the first step, but that could have been lengthier and more complex to solve. Also, be careful about the signs in the formulas that we used and what is asked in the question. It is also necessary that you learn all the basic algebraic formulas and binomial expansions as they are often used in solving problems.