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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 $180$.
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