Find the value of $x$, if:
(i) $4x = (52)^2 – (48)^2$
(ii) $14x = (47)^2 – (33)^2$
(iii) $5x = (50)^2 – (40)^2$

Given:

(i) $4x = (52)^2 – (48)^2$ 

(ii) $14x = (47)^2 – (33)^2$ 

(iii) $5x = (50)^2 – (40)^2$

To do:

We have to find the value of $x$ in each case.

Solution:

Here, we have to find the value of $x$ in each expression. The given expressions are the difference of two squares. So, to find the value of $x$ we can simplify the RHS in each case using the identity:

$(a – b) (a + b) = a^2 – b^2$.

Therefore,

(i) $4x = (52)^2 – (48)^2$ 

This implies,

$4x=(52+48)\times(52-48)$

$4x=100\times4$

$4x=400$

$x=\frac{400}{4}$

$x=100$

Hence, the value of $x$ is $100$.

(ii) $14x = (47)^2 – (33)^2$ 

This implies,

$14x=(47+33)\times(47-33)$

$14x=80\times14$

$x=\frac{80\times14}{14}$

$x=80$

Hence, the value of $x$ is $80$.

(iii) $5x = (50)^2 – (40)^2$

This implies,

$5x=(50+40)\times(50-40)$

$5x=90\times10$

$x=\frac{90\times10}{5}$

$x=90\times2$

$x=180$

Hence, the value of $x$ is $180$.