Question

Find x, if
$x(x+7)=\frac{(x^2+11x+28)}{(x+4)}.$

• A. 0.0
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is C 2

Given, the expression is
$x(x+7)=\frac{(x^2+11x+28)}{(x+4)}...\left(i\right)$
Now, Comparing the expression $x^2+11x+28$ with the identity $x^2+(a+b)x+ab$
$\text{we note that,}$
$(a+b)=11andab=28$
So,
$(7+4)=11and\left(7\right)\left(4\right)=28$
Hence,
$x^2+11x+28$
$=x^2+7x+4x+28$
$=x(x+7)+4(x+7)$
$=(x+4)(x+7)$

Now, from (i)
$x(x+7)=\frac{(x^2+11x+28)}{(x+4)}$
$\Rightarrow x(x+7)=\frac{(x+4)(x+7)}{(x+4)}$
After canceling the common terms, we get $x=1.$