Question

1. Which domain restrictions apply to the rational expression?

Answer 1

for question #1 the answer is: x=2, x=-1

question 9: answer is B

question 10: answer B

Answer 2

Answer with explanation:

1. The given rational expression is

   $\rightarrow\frac{x^2+5 x+6}{x^2-9}\\\\\rightarrow\frac{x^2+5 x+6}{(x-3)(x+3)}$

The function is not defined ,when

 →(x-3)(x+3)=0

→x-3≠0 ∧ x+3≠0

→x≠3, ∧ x ≠ -3

⇒Option C and D

→x≠−3

→x≠3

2.

$\rightarrow\frac{4 x}{x-3}+\frac{2}{x^2-9}=\frac{1}{x+3}\\\\\rightarrow\frac{4 x}{x-3}+\frac{2}{(x-3)(x+3)}=\frac{1}{x+3}\\\\\rightarrow\frac{4 x(x+3)+2}{(x-3)(x+3)}=\frac{1}{x+3}\\\\\rightarrow4 x(x+3)+2=\frac{(x-3)(x+3)}{x+3}\\\\\rightarrow4x^2+12 x+2=x-3\\\\\rightarrow4x^2+11x+5=0\\\\ \text{Using Discriminant method for a quadratic equation}\\\\ax^2+bx +c=0\\\\x=\frac{-b\pm\sqrt{D}}{2 a}\\\\D=b^2-4 ac\\\\x=\frac{-11\pm\sqrt{121-80}}{2 \times 4}\\\\x=\frac{-11\pm\sqrt{41}}{8}$

None of the option

3.

$\rightarrow \frac{1}{x}+\frac{1}{x-3}=\frac{x-2}{x-3}\\\\\rightarrow\frac{x-3+x}{x(x-3)}=\frac{x-2}{x-3}\\\\\rightarrow2x-3=\frac{x(x-3)(x-2)}{x-3}\\\\\rightarrow 2 x-3=x^2-2 x\\\\\rightarrow x^2-4x+3=0\\\\\rightarrow (x-1)(x-3)=0\\\\x=1,3$

For, x=3 , the equation is not defined.

So, there is single solution which is , x=1.

Option B:→ There is only one solution: x = 1.

The solution x = 3 is an extraneous solution.

4.

$\rightarrow \sqrt{3}x+1-x+3=0\\\\\rightarrow \sqrt{3}x -x=-4\\\\\rightarrow x(\sqrt{3}-1)=-4\\\\\rightarrow x=\frac{-4}{\sqrt{3}-1}\\\\\rightarrow x=\frac{-4\times(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}\\\\x=\frac{-4\times(\sqrt{3}+1)}{2}\\\\x=-2(\sqrt{3}+1)$

None of the option