Question

which of the following is correct ?

Answer 1

The product of the additive inverse of$\frac{x^2-9}{2+x}$ and the additive inverse of $\frac{x^2-4}{x-3}$ is $x^2+x-6$ (Option b)

Given

Two quadratic equations, $\frac{x^2-9}{2+x}$ and $\frac{x^2-4}{x-3}$

To Find

Product of additive inverse of the equations

Solution

The additive inverse of $\frac{x^2-9}{2+x}$  is $-(\frac{x^2-9}{2+x})$ which can be represented as $\frac{9-x^2}{2+x}$

The additive inverse of $\frac{x^2-4}{x-3}$ is $-(\frac{x^2-4}{x-3})$ which can be represented as $\frac{4-x^2}{x-3}$

Multiplying the additive inverses we get,

$(\frac{9-x^2}{2+x})$$(\frac{4-x^2}{x-3})$ = $(\frac{(3+x)(3-x)}{(x+2)})$ $(\frac{(2+x)(2-x)}{x-3} )$ =$\frac{(3+x)(3-x)(2+x)(2-x))}{(x+2)(x-3)}$ = $(3+x)(-1)(2-x)$ = $(x+3)(x-2)$ = $x^2-2x+3x-6$ = $x^2+x-6$

Hence, the product of the additive inverse of$\frac{x^2-9}{2+x}$ and the additive inverse of $\frac{x^2-4}{x-3}$ is $x^2+x-6$

Answer 2

Answer:

The correct answer is option(b)

Step-by-step explanation:

Recall the concept

Two expressions are said to be additive inverse of the other if the sum of the two expressions is zero

a²-b² = (a+b)(a-b)

Additive inverse of $\frac{x^{2} - 9}{2+x}$ = $\frac{9 - x^2}{2+x}$ (since  $\frac{x^{2} - 9}{2+x}$+ $\frac{x^{2} - 9}{2+x}$ = 0)

Additive inverse of $\frac{x^{2} - 4}{x-3}$ = $\frac{4 - x^2}{x-3}$ (  since  $\frac{x^{2} - 4}{x-3}$+ $\frac{4 - x^2}{x-3}$ = 0)

Required to find the product of additive inverses of  $\frac{x^{2} - 9}{2+x}$ and  $\frac{x^{2} - 4}{x-3}$

= $\frac{9 - x^2}{2+x}$× $\frac{4-x^2}{x-3}$

Applying the identity a²-b² = (a+b)(a-b) we get

$\frac{9 - x^2}{2+x}$× $\frac{4 -x^2}{x-3}$ = $\frac{(3+x)(3-x)(2+x)(2-x)}{(2+x)(x-3)}$

= -(3+x)(2-x)

=(x+3)(x-2)

= x² +3x -2x -6

= x² +x -6

The product of additive inverses of  $\frac{x^{2} - 9}{2+x}$ and  $\frac{x^{2} - 4}{x-3}$  = x² +x -6

The correct answer is option (b)

#SPJ6