36. For what values of n will 𝑎2 + 𝑛 = 3𝑎𝑛 − 1 −
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
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 $\boxed{\textbf{(D) }n=0,\; n=-\frac{7}{9}}$.
To determine the values of n that will result in repeated roots for the given equation, we can use the discriminant of the quadratic equation.
The given equation is:
𝑎^2 + 𝑛 = 3𝑎𝑛 - 1 - 2𝑎
To put it in the standard quadratic form, we move all terms to one side:
𝑎^2 - 3𝑎𝑛 + 𝑛 + 2𝑎 + 1 = 0
Now, we can compare this equation with the standard quadratic equation 𝑎^2 + 𝑏𝑎 + 𝑐 = 0.
From the comparison, we can see that:
𝑏 = -3𝑛 + 2
𝑐 = 𝑛 + 1
The discriminant, D, of a quadratic equation 𝑎^2 + 𝑏𝑎 + 𝑐 = 0 is given by:
D = 𝑏^2 - 4𝑎𝑐
For repeated roots, the discriminant should be equal to zero.
So, setting D = 0, we have:
(-3𝑛 + 2)^2 - 4(1)(𝑛 + 1) = 0
Expanding and simplifying:
9𝑛^2 - 12𝑛 + 4 - 4𝑛 - 4 = 0
9𝑛^2 - 16𝑛 = 0
Factoring:
𝑛(9𝑛 - 16) = 0
Setting each factor equal to zero, we have two possible values for n:
1) 𝑛 = 0
2) 9𝑛 - 16 = 0 => 9𝑛 = 16 => 𝑛 = 16/9
Therefore, the values of n that will result in repeated roots are:
A. 𝑛 = 0, 𝑛 = 16/9
So, the correct answer is not given in the options provided.