Respuesta :
Answer:
[tex]a=0\\a=2\\a=-2[/tex]
Step-by-step explanation:
We have [tex]\frac{3}{a+2}+\frac{2}{a} =\frac{4a-4}{a^2-4}[/tex]
We are going to analyze both sides of the equation separately.
Part a: [tex]\frac{3}{a+2}+\frac{2}{a}[/tex]
Applying common denominator:
[tex]\frac{3}{a+2}+\frac{2}{a}=\frac{3a+2(a+2)}{a(a+2)}[/tex]
[tex]\frac{3a+2(a+2)}{a(a+2)}=\frac{3a+2a+4}{a(a+2)}=\frac{5a+4}{a(a+2)}[/tex]
Part b:
[tex]\frac{4a-4}{a^2-4}[/tex]
In the numerator we can apply common factor 4. And in the denominator we can apply difference of squares.
Remember: Difference of squares: [tex](a^2-b^2)=(a+b)(a-b)[/tex]
[tex]\frac{4a-4}{a^2-4}=\frac{4(a-1)}{a^2-2^2}=\frac{4(a-1)}{(a+2)(a-2)}[/tex]
Then,
[tex]\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)}[/tex]
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:
[tex]a(a+2)[/tex] and we have to see for which values of [tex]a[/tex] the denominator is zero:
[tex]a(a+2)=0[/tex] ⇒ [tex]a=0[/tex] and [tex]a=-2[/tex]
The denominator of the second side of the equation is:
[tex](a+2)(a-2)[/tex]
And we have to see for which values of [tex]a[/tex] the expression is zero,
[tex](a+2)(a-2)=0[/tex] ⇒ [tex]a=-2[/tex] and [tex]a=2[/tex]
Then the extraneous solutions of the equation are:
[tex]a=0\\a=2\\a=-2[/tex]
Because those are the values of [tex]a[/tex] that make the denominator zero.
Answer:
Solutions : x=4,-2
Extraneous solution : x=-2.
Step-by-step explanation:
The given equation is
[tex]\dfrac{3}{a+2}+\dfrac{2}{a}=\dfrac{4a-4}{a^2-4}[/tex]
Taking LCM we get
[tex]\dfrac{3a+2(a+2)}{(a+2)a}=\dfrac{4a-4}{(a-2)(a+2)}[/tex] [tex][\because a^2-b^2=(a-b)(a+b)][/tex]
[tex]\dfrac{3a+2a+4}{(a+2)a}=\dfrac{4a-4}{(a-2)(a+2)}[/tex]
Cancel out common factors from the denominators.
[tex]\dfrac{5a+4}{a}=\dfrac{4a-4}{a-2}[/tex]
On cross multiplication we get
[tex](5a+4)(a-2)=(4a-4)a[/tex]
[tex]5a^2-10a+4a-8=4a^2-4a[/tex]
[tex]5a^2-6a-8-4a^2+4a=0[/tex]
[tex]a^2-2a-8=0[/tex]
Splitting the middle term we get
[tex]a^2-4a+2a-8=0[/tex]
[tex]a(a-4)+2(a-4)=0[/tex]
[tex](a+2)(a-4)=0[/tex]
Using zero product property we get
[tex]a=-2,a=4[/tex]
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.
