36. For what values of n will 𝑎2 + 𝑛 = 3𝑎𝑛 − 1 −

Question

2𝑎 have repeated roots?
A. 𝑛 = 0, 𝑛 = 1 7
9
.
B. 𝑛 = 0, 𝑛 = 4
C. 𝑛 = 0, 𝑛 = −1 1
2
D. 𝑛 = 0, 𝑛 = −1 7
9
3

Answer 1

We can rewrite the equation as a quadratic in terms of $a$: $$a^2 - (3n)a +(n+2) = 0.$$ For the roots of this quadratic to be equal, the discriminant must be equal to zero: $$(-3n)^2 - 4(n+2) = 0.$$ Solving for $n$, we get $n=0$ or $n= -\frac{7}{9}$. Therefore, the answer is $\boxed{\textbf{(D) }n=0,\; n=-\frac{7}{9}}$.