Question

How do you solve $4{x^2} = 81$?

Answer 1

Hint: In this question, we need to solve the given equation. Note that the given equation is a quadratic equation. We solve this by finding the value of the variable x. Firstly we will move all the coefficient terms in the left hand side to the right hand side by dividing the both sides of the equation by 4. Then we take the square root on both sides. After that we simplify the expression to get the value for the variable x. Here we get two solutions for the given equation.

Complete step by step answer:
Given the equation $4{x^2} = 81$ …… (1)
We are asked to solve the above expression given in the equation (1). i.e. we need to find the value of the unknown variable x.
Note that the above equation is a quadratic equation of degree two.
We will simplify the given equation and solve for x.
Firstly, let us move all the coefficients on the left hand side to the right hand side.
We do this, by dividing each of the sides of the equation (1) by 4, we get,
$ \Rightarrow \dfrac{{4{x^2}}}{4} = \dfrac{{81}}{4}$
Cancelling the common terms and simplifying we get,
$ \Rightarrow {x^2} = \dfrac{{81}}{4}$
Now taking square root on both sides we get,
$ \Rightarrow \sqrt {{x^2}} = \pm \sqrt {\dfrac{{81}}{4}} $
This can also be written as,
$ \Rightarrow \sqrt {{x^2}} = \pm \dfrac{{\sqrt {81} }}{{\sqrt 4 }}$
We know that the value of $\sqrt {81} = 9$ and $\sqrt 4 = 2$.
Hence substituting these values we get,
$ \Rightarrow \sqrt {{x^2}} = \pm \dfrac{9}{2}$
$ \Rightarrow x = \pm \dfrac{9}{2}$

Thus, the solution for the given equation $4{x^2} = 81$ is given by $x = \pm \dfrac{9}{2}$.

Note: Alternative method :
Given the equation $4{x^2} = 81$
Transferring the constant term on the R.H.S. which is 81 to the L.H.S. we get,
$ \Rightarrow 4{x^2} - 81 = 0$
We can write $4{x^2} = {(2x)^2}$ and $81 = {9^2}$.
Hence the above expression can be written as,
$ \Rightarrow {(2x)^2} - {9^2} = 0$
This is of the form ${a^2} - {b^2}$.
We have the algebraic formula given by, ${a^2} - {b^2} = (a - b)(a + b)$
Here $a = 2x$ and $b = 9$.
Hence applying the formula, we get,
$ \Rightarrow (2x - 9)(2x + 9) = 0$
$ \Rightarrow 2x - 9 = 0$ or $2x + 9 = 0$
If $2x - 9 = 0$, then we have,
$ \Rightarrow 2x = 9$
$ \Rightarrow x = \dfrac{9}{2}$
If $2x + 9 = 0$, then we have,
$ \Rightarrow 2x = - 9$
$ \Rightarrow x = - \dfrac{9}{2}$
Therefore we have $x = \pm \dfrac{9}{2}$
Hence we get the solution for the given equation as $x = \pm \dfrac{9}{2}$.