Question

If $\alpha$, $\alpha + 1$ are the roots of the equation $x^2 + nx + (n + 5) = 0$ then the non-negative value of n is equal to

• A. $\alpha^2 + 2$
• B. $2 - 3\alpha$
• C. $3 - \alpha$
• D. $\alpha^2 - 2$

Answer 1

Answer: (C)

$3 - \alpha$

Answer 2

The given quadratic equation is $x^2 + nx + (n + 5) = 0$. Let the roots be $r_1 = \alpha$ and $r_2 = \alpha + 1$.

According to Vieta's formulas:

Sum of roots: $r_1 + r_2 = -\frac{n}{1}$

$\alpha + (\alpha + 1) = -n$

$2\alpha + 1 = -n$

$n = -(2\alpha + 1)$ (Equation 1)

Product of roots: $r_1 \cdot r_2 = \frac{n + 5}{1}$

$\alpha(\alpha + 1) = n + 5$

$\alpha^2 + \alpha = n + 5$ (Equation 2)

Substitute Equation 1 into Equation 2:

$\alpha^2 + \alpha = -(2\alpha + 1) + 5$

$\alpha^2 + \alpha = -2\alpha - 1 + 5$

$\alpha^2 + \alpha = -2\alpha + 4$

$\alpha^2 + 3\alpha - 4 = 0$

Solve the quadratic equation for $\alpha$:

$(\alpha + 4)(\alpha - 1) = 0$

This gives two possible values for $\alpha$:

$\alpha = -4$ or $\alpha = 1$.

Now, find the corresponding value of $n$ for each value of $\alpha$ using Equation 1 ($n = -(2\alpha + 1)$).

Case 1: $\alpha = -4$

$n = -(2(-4) + 1) = -(-8 + 1) = -(-7) = 7$. In this case, the roots are $\alpha = -4$ and $\alpha + 1 = -3$. The equation is $x^2 + 7x + (7+5) = 0 \implies x^2 + 7x + 12 = 0$. The roots of $x^2+7x+12=0$ are indeed $-3$ and $-4$. The value of $n$ is 7, which is non-negative.

Case 2: $\alpha = 1$

$n = -(2(1) + 1) = -(2 + 1) = -3$. In this case, the roots are $\alpha = 1$ and $\alpha + 1 = 2$. The equation is $x^2 - 3x + (-3+5) = 0 \implies x^2 - 3x + 2 = 0$. The roots of $x^2-3x+2=0$ are indeed $1$ and $2$. The value of $n$ is -3, which is negative.

The question asks for the non-negative value of $n$. From the two cases, the non-negative value of $n$ is 7. This non-negative value of $n=7$ occurs when $\alpha = -4$.

Now we need to find which of the given options is equal to 7, using the value of $\alpha$ that corresponds to $n=7$, which is $\alpha = -4$.

Let's evaluate each option with $\alpha = -4$:

(1) $\alpha^2 + 2 = (-4)^2 + 2 = 16 + 2 = 18$.

(2) $2 - 3\alpha = 2 - 3(-4) = 2 + 12 = 14$.

(3) $3 - \alpha = 3 - (-4) = 3 + 4 = 7$.

(4) $\alpha^2 - 2 = (-4)^2 - 2 = 16 - 2 = 14$.

Option (3) evaluates to 7, which is the non-negative value of $n$.