Question

Which solution to the equation (3/a+2)+2/a = 4a-4/a^2-4 is extraneous

Answer 1

Answer:

Solutions : x=4,-2

Extraneous  solution : x=-2.

Step-by-step explanation:

The given equation is

$\dfrac{3}{a+2}+\dfrac{2}{a}=\dfrac{4a-4}{a^2-4}$

Taking LCM we get

$\dfrac{3a+2(a+2)}{(a+2)a}=\dfrac{4a-4}{(a-2)(a+2)}$         $[\because a^2-b^2=(a-b)(a+b)]$

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

Cancel out common factors from the denominators.

$\dfrac{5a+4}{a}=\dfrac{4a-4}{a-2}$

On cross multiplication we get

$(5a+4)(a-2)=(4a-4)a$

$5a^2-10a+4a-8=4a^2-4a$

$5a^2-6a-8-4a^2+4a=0$

$a^2-2a-8=0$

Splitting the middle term we get

$a^2-4a+2a-8=0$

$a(a-4)+2(a-4)=0$

$(a+2)(a-4)=0$

Using zero product property we get

$a=-2,a=4$

Extraneous solutions: From the solutions of an equation, the invalid solutions are known as extraneous solutions.

For a=-2 right hand side of the given equation is not defined because the denominator become 0.

Therefore, -2 is an extraneous solution.

Answer 2

Answer:

$a=0\\a=2\\a=-2$

Step-by-step explanation:

We have $\frac{3}{a+2}+\frac{2}{a} =\frac{4a-4}{a^2-4}$

We are going to analyze  both sides of the equation separately.

Part a: $\frac{3}{a+2}+\frac{2}{a}$

Applying common denominator:

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

$\frac{3a+2(a+2)}{a(a+2)}=\frac{3a+2a+4}{a(a+2)}=\frac{5a+4}{a(a+2)}$

Part b:

$\frac{4a-4}{a^2-4}$

In the numerator we can apply common factor 4. And in the denominator we can apply difference of squares.

Remember: Difference of squares: $(a^2-b^2)=(a+b)(a-b)$

$\frac{4a-4}{a^2-4}=\frac{4(a-1)}{a^2-2^2}=\frac{4(a-1)}{(a+2)(a-2)}$

Then,

$\frac{3}{a+2}+\frac{2}{a} =\frac{4a-4}{a^2-4}\\\\\frac{5a+4}{a(a+2)}=\frac{4(a-1)}{(a+2)(a-2)}$

An extraneous solution is not a valid solution for a problem. We know that the denominator can't be zero.

The denominator of the first side of the equation is:

$a(a+2)$ and we have to see for which values of $a$ the denominator is zero:

$a(a+2)=0$ ⇒ $a=0$ and $a=-2$

The denominator of the second side of the equation is:

$(a+2)(a-2)$

And we have to see for which values of $a$ the expression is zero,

$(a+2)(a-2)=0$ ⇒ $a=-2$ and $a=2$

Then the extraneous solutions of the equation are:

$a=0\\a=2\\a=-2$

Because those are the values of $a$  ​​that make the denominator zero.