Question

The positive square root of
$\frac{{(a+b)}^2-{(c+d)}^2}{{(a+b)}^2-{(c-d)}^2}\times \frac{{(a+b+c)}^2-d^2}{{(a+b-c)}^2-d^2}$ using factorisation is:

• A. $\frac{a+b+c+d}{a+b-c+d}$
• B. $\frac{a+b+c-d}{a+b+c+d}$
• C. $\frac{a+b-c+d}{a+b+c-d}$
• D. $\frac{a-b-c+d}{a+b+c+d}$

Answer 1

Answer: (A)

$\frac{a+b+c+d}{a+b-c+d}$

$\frac{{(a+b)}^2-{(c+d)}^2}{{(a+b)}^2-{(c-d)}^2}\times \frac{{(a+b+c)}^2-d^2}{{(a+b-c)}^2-d^2}$

$=\frac{((a+b)+(c+d))((a+b)-(c+d))}{((a+b)-(c-d))((a+b)+(c-d))}\times \frac{(a+b+c-d)(a+b+c+d)}{(a+b-c-d)(a+b-c+d)}$

$=\frac{(a+b+c+d)(a+b-c-d)}{(a+b-c+d)(a+b+c-d)}\times \frac{(a+b+c-d)(a+b+c+d)}{(a+b-c-d)(a+b-c+d)}$

$=[\frac{a+b+c+d}{a+b-c+d}{]}^2$

Square root is,
$\frac{a+b+c+d}{a+b-c+d}$