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Goryan [66]
3 years ago
13

Solve equation, check for extraneous solutions

Mathematics
1 answer:
fiasKO [112]3 years ago
8 0

Start with

1=\dfrac{1}{n^2}+\dfrac{n^2-4n-5}{3n^2}

We observe that both fractions are not defined if n=0. So, we will assume n \neq 0.

We multiply both numerator and denominator of the first fraction by 3 and we sum the two fractions:

1=\dfrac{3}{3n^2}+\dfrac{n^2-4n-5}{3n^2} = \dfrac{n^2-4n-2}{3n^2}

We multiply both sides by 3n^2:

n^2-4n-2=3n^2

We move everything to one side and solve the quadratic equation:

2n^2+4n+2=0 \iff 2(n^2+2n+1)=0 \iff 2(n+1)^2=0 \iff n+1=0 \iff n=-1

We check the solution:

1=\dfrac{1}{1}+\dfrac{1+4-5}{3} \iff 1 = 1+0

which is true

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