Question

A student solves the following equation and
determines that the solution is -2. Is the student
correct? Explain.
3/a+2 - 6a/a^2-4 = 1/a-2

Answer 1

Answer:

Since the equation is undefined for -2

Therefore, NO SOLUTION for the given equation.

Step-by-step explanation:

Considering the expression

$\frac{3}{a+2}-6\cdot \frac{a}{-4+a^2}=\frac{1}{a-2}$

$\frac{3}{a+2}-\frac{6a}{-4+a^2}=\frac{1}{a-2}$

$\mathrm{Find\:Least\:Common\:Multiplier\:of\:}a+2,\:-4+a^2,\:a-2:\quad \left(a+2\right)\left(a-2\right)$

$\mathrm{Multiply\:by\:LCM=}\left(a+2\right)\left(a-2\right)$

$\frac{3}{a+2}\left(a+2\right)\left(a-2\right)-\frac{6a}{-4+a^2}\left(a+2\right)\left(a-2\right)=\frac{1}{a-2}\left(a+2\right)\left(a-2\right)$

as

• $\frac{3}{a+2}\left(a+2\right)\left(a-2\right):\quad 3\left(a-2\right)$

• $-\frac{6a}{-4+a^2}\left(a+2\right)\left(a-2\right):\quad -6a$

• $\frac{1}{a-2}\left(a+2\right)\left(a-2\right):\quad a+2$

so equation becomes

$3\left(a-2\right)-6a=a+2$  

$-3a-6=a+2$

$-3a-6+6=a+2+6$

$-4a=8$

$\mathrm{Divide\:both\:sides\:by\:}-4$

$\frac{-4a}{-4}=\frac{8}{-4}$

$a=-2$

$\mathrm{Verify\:Solutions}$

$\mathrm{Take\:the\:denominator\left(s\right)\:of\:}\frac{3}{a+2}-6\frac{a}{-4+a^2}-\frac{1}{a-2}\mathrm{\:and\:compare\:to\:zero}$

$\mathrm{Solve\:}\:a+2=0:\quad a=-2$

$\mathrm{Solve\:}\:-4+a^2=0:\quad a=2,\:a=-2$

$\mathrm{Solve\:}\:a-2=0:\quad a=2$

So the following points are undefined

$a=-2,\:a=2$

Since the equation is undefined for -2

Therefore, NO SOLUTION for the given equation.

Answer 2

Solving the equation, it is found that it has no solutions, thus, the student is not correct.

The equation is:

$\frac{3}{a + 2} - \frac{6a}{a^2 - 4} = \frac{1}{a - 2}$

Applying the subtraction of perfect squares, we have that:

$a^2 - 4 = (a - 2)(a + 2)$

Then

$\frac{3}{a + 2} - \frac{6a}{(a - 2)(a + 2)} = \frac{1}{a - 2}$

$\frac{3}{a + 2} - \frac{6a}{(a - 2)(a + 2)} - \frac{1}{a - 2} = 0$

$\frac{3(a - 2) - 6a - (a + 2)}{(a - 2)(a + 2)} = 0$

$\frac{3a - 6 - 6a - a - 2}{(a - 2)(a + 2)} = 0$

$\frac{-4a - 8}{(a - 2)(a + 2)} = 0$

In a fraction, the denominator cannot be zero, thus:

$a - 2 \neq 0 \rightarrow a \neq 2$

$a + 2 \neq 0 \rightarrow a \neq -2$

Looking at the numerator for the solution:

$-4a - 8 = 0$

$4a = -8$

$a = -\frac{8}{4}$

$a = -2$

However, a = -2 is a restriction, as it makes the denominator zero, thus the function has no solutions.

A similar problem is given at https://brainly.com/question/18734659