Question

How do you solve $9{n^2} + 10 = 91$?

Answer 1

The value of n comes out to be 3 and -3.
Try to solve the question by completing the square method by transforming the given equation in the form of ${(a + b)^2} = 0$ by adding or subtracting something from both sides of the equation.

Complete step-by-step solution-
Given equation,
$9{n^2} + 10 = 91$
The first step is to make the left-hand side of the equation equal to zero.
For this, we add ‘-91’ to both sides of the equation.
After performing the calculation, we get
$\Rightarrow 9{n^2} + 10 - 91 = 91 - 91$
$\Rightarrow 9{n^2} - 81 = 0$
To apply to convert the square method we have to transform the equation in the form of ${x^2} + bx + c = 0$ (notice carefully, here the coefficient of the term ${x^2}$ is unity) but in the given equation, the coefficient of ${x^2}$ is 9.
Therefore we need to divide the given equation by 9 to make the coefficient of ${x^2}$ unity.
After dividing by 9 into both sides, we get
$\Rightarrow \dfrac{{9{n^2} - 81}}{9} = 0$
On further simplifying, we get
$\Rightarrow \dfrac{{9{n^2}}}{9} - \dfrac{{81}}{8} = 0$
$\Rightarrow {n^2} - 9 = 0$
Bring the constant term to the left-hand side by adding 9 to both sides of the equation.
$\Rightarrow {n^2} = 9$
To obtain the final answer we need to simplify the equation further by taking the square root on both sides.
Hence, after taking the square root on both sides,
$\Rightarrow \sqrt {{n^2}} = \sqrt 9$
As we know that $\sqrt 9 = \pm 3$, we get
$\Rightarrow n = \pm 3$
Which is the required solution of the equation $9{n^2} + 10 = 91$.

Note- The above question can also be solved by various other methods but if any specific method is not mentioned in the question you should opt for the above-discussed method. There are two values of n as the equation $9{n^2} + 10 = 91$ is a quadratic equation and will have two roots. You can also verify your solution by substituting the calculated values of n in the equation. If the value satisfies the equation your answer is correct as roots of any equation are also the solutions of that particular equation.