Question

1. Which domain restrictions apply to the rational expression?
x^2+5x+6
--------------
x^2-9

Select each correct answer.
x≠2
x≠1
x≠−3
x≠3
x≠−2
x≠−1


(4x)/(x-3)+(2)/(x^(2)-9)= (1)/(x+3)
A. 19/3
B. 13/3
C. -5
D. -17/3

9. Which statement best reflects the solution(s) of the equation?
1/x + 1/x-3 = x-2 / x-3
A. There is only one solution: x = 1.
The solution x = 0 is an extraneous solution.

B. There is only one solution: x = 1.
The solution x = 3 is an extraneous solution.

C. There is only one solution: x = 3.
The solution x = 1 is an extraneous solution.

D. There are two solutions: x = 1 and x = 3.

10. (√3x+1)−x+3=0
A. There is only one solution: x = 1.
The solution x = 8 is an extraneous solution.

B. There is only one solution: x = 8.
The solution x = 1 is an extraneous solution.

C. There is only one solution: x = 8.
The solution x = 0 is an extraneous solution.

D. There are two solutions: x = 1 and x = 8.

Answer 1

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

Answer 2

1) Options c) and d) are correct.

2) None of the options are correct.

3) Option B) is correct.

4) None of the options are correct.

Step-by-step explanation:

1) Given :   $4x^2+12x+2=x-3$

  Expression --         $\dfrac{x^2+5x+6}{x^2-9}$

  Solution :

  $\dfrac{x^2+5x+6}{x^2-9}=\dfrac{x^2+3x+2x+6}{(x+3)(x-3)}$

  $\dfrac{x^2+5x+6}{x^2-9}=\dfrac{(x+3)(x+2)}{(x+3)(x-3)}$

  Therefore, $\rm x\neq 3 \; and\;x\neq -3$ ,option c) and d) is correct.

2) Given :

    Expression - $\dfrac{4x}{x-3}+\dfrac{2}{x^2-9}=\dfrac{1}{x+3}$

    Solution :

     $\dfrac{4x}{(x-3)}+\dfrac{2}{(x+3)(x-3)}=\dfrac{1}{(x+3)}$

     $\dfrac{4x(x+3)+2}{(x-3)(x+3)}=\dfrac{1}{(x+3)}$

     $4x^2+12x+2=x-3$

     $4x^2+11x+5=0$

     $x=\dfrac{-11\pm\sqrt{121-80} }{8}$

     $x = \dfrac{-11\pm\sqrt{41} }{8}$

    None of the options are correct.

3) Given :

  Expression -     $\dfrac{1}{x}+\dfrac{1}{x-3}=\dfrac{x-2}{x-3}$     ----- (1)

   Solution :

   $\dfrac{x-3+x}{(x)(x-3)}=\dfrac{x-2}{x-3}$

   $2x-3=x(x-2)$

    $x^2-4x+3=0$

    $x^2-3x-x+3=0$

    $(x-3)(x-1)=0$

At x = 3 equation (1) is not define. Therefore, the correct answer is option

B) There is only one solution: x = 1. The solution x = 3 is an extraneous solution.

4) Given :

   Exprression -  $(\sqrt{3}x +1)-x+3=0$

   Solution :

   $x(\sqrt{3}-1 )=-4$

   $x=\dfrac{-4}{\sqrt{3}-1 }$

   None of the options are correct.

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