Question

If $x + \dfrac{9}{x} = 6$, then ${x^2} + \dfrac{9}{{{x^2}}} = $?

Answer 1

Hint: Here, in the given question, we are given $x + \dfrac{9}{x} = 6$ and we need to find the value of ${x^2} + \dfrac{9}{{{x^2}}}$. At first we will find the value of $x$, using $x + \dfrac{9}{x} = 6$. To find the value of $x$, we will convert $x + \dfrac{9}{x} = 6$ into a quadratic equation. By using the factorization method we will find the value of $x$. In this method, we express $a{x^2} + bx + c = 0$ as the product of two linear factors, say $\left( {px + q} \right)$ and $\left( {rx + s} \right)$, where $p$, $q$, $r$ and $s$ are real numbers such that $p \ne 0$ and $r \ne 0$.

Complete step-by-step answer:
Given that, $x + \dfrac{9}{x} = 6$
Take LCM on this left-hand side
$ \Rightarrow \dfrac{{x \times x + 9}}{x} = 6$
$ \Rightarrow \dfrac{{{x^2} + 9}}{x} = 6$
On cross-multiplication, we get
$ \Rightarrow {x^2} + 9 = 6x$
On shifting $6x$ to left-hand side, we get
$ \Rightarrow {x^2} - 6x + 9 = 0$
Let us express ${x^2} - 6x + 9 = 0$ as the product of two linear factors.
$ \Rightarrow {x^2} - 3x - 3x + 9 = 0$
On taking common factors, we get
$ \Rightarrow x\left( {x - 3} \right) - 3\left( {x - 3} \right) = 0$
$ \Rightarrow \left( {x - 3} \right)\left( {x - 3} \right) = 0$
The product of multiplication of exponents with the same base is equal to the sum of their powers with the same base. Therefore, we get
$ \Rightarrow {\left( {x - 3} \right)^2} = 0$
On taking square root on both sides, we get
$ \Rightarrow x - 3 = 0$
$ \Rightarrow x = 3$
We need to find the value of ${x^2} + \dfrac{9}{{{x^2}}}$. On substituting the value of $x$, we get
$ \Rightarrow {3^2} + \dfrac{9}{{{3^2}}}$
$ \Rightarrow 9 + \dfrac{9}{9}$
On cancelling out common factors, we get
$ \Rightarrow 9 + 1$
$ \Rightarrow 10$
Therefore, the value of ${x^2} + \dfrac{9}{{{x^2}}}$ is $10$.
So, the correct answer is “10”.

Note: A quadratic equation of the form $a{x^2} + bx + c = 0$ where $a \ne 0$ can be solved by many methods. Such as, completing the square, factorization method, by using quadratic formula, etc. Doesn’t matter with which method we solve any quadratic equation because the roots of the quadratic equation will remain the same.