Question

Let P(x) = $x^3$ + $ax^2$ + bx be a polynomial whose roots are non-negative and are in arithmetic Progression. If the sum of coefficients of P(x) is 10, then:

• A. sum of the roots of P(x) is equal to 9
• B. sum of the roots of P(x) is equal to 18
• C. The value of (b – a) is equal to 9
• D. The value of (b – a) is equal to 27

Answer 1

Answer: (A), (D)

A) sum of the roots of P(x) is equal to 9

D) The value of (b – a) is equal to 27

Answer 2

Let the polynomial be $P(x) = x^3 + ax^2 + bx$. Let the roots of the polynomial be $\alpha, \beta, \gamma$. According to Vieta's formulas:

• Sum of the roots: $\alpha + \beta + \gamma = -a$
• Sum of the products of the roots taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = b$
• Product of the roots: $\alpha\beta\gamma = -0 = 0$

The condition that the product of the roots is 0 implies that at least one of the roots is 0. The roots are non-negative and are in arithmetic progression (AP). Let the roots be $k-d, k, k+d$. Since the roots are non-negative, we have $k-d \ge 0$, $k \ge 0$, and $k+d \ge 0$. Since at least one root is 0, one of $k-d, k, k+d$ must be 0.

Case 1: $k-d = 0$. This implies $k=d$. The roots are $0, d, 2d$. Since the roots are non-negative, we must have $d \ge 0$. If $d=0$, the roots are $0, 0, 0$. If $d>0$, the roots are $0, d, 2d$, which are distinct non-negative numbers in AP.

Case 2: $k = 0$. The roots are $-d, 0, d$. Since the roots are non-negative, we must have $-d \ge 0$ and $d \ge 0$. This implies $d=0$. The roots are $0, 0, 0$.

Case 3: $k+d = 0$. Since $k \ge 0$ and $d \ge 0$ (because $k-d \ge 0$ and $k \ge 0$ implies $d \le k$, so if $d<0$, $k-d>k\ge 0$, but $k+d=0$ implies $d=-k \le 0$. Hence $d$ must be 0, which implies $k=0$), this is only possible if $k=0$ and $d=0$. The roots are $0, 0, 0$.

So, the roots are either $0, 0, 0$ (when $d=0$) or $0, d, 2d$ with $d>0$.

The sum of the coefficients of P(x) is given as 10. The sum of coefficients of a polynomial $P(x)$ is $P(1)$. $P(1) = 1^3 + a(1)^2 + b(1) = 1 + a + b$. Given $1 + a + b = 10$, so $a + b = 9$.

Let's consider the roots $0, d, 2d$. Sum of roots = $0 + d + 2d = 3d$. From Vieta's formulas, sum of roots = $-a$. So, $3d = -a$, which means $a = -3d$.

Sum of products of roots taken two at a time = $(0)(d) + (d)(2d) + (2d)(0) = 0 + 2d^2 + 0 = 2d^2$. From Vieta's formulas, sum of products of roots taken two at a time = $b$. So, $b = 2d^2$.

Substitute the expressions for $a$ and $b$ into the equation $a + b = 9$: $(-3d) + (2d^2) = 9$ $2d^2 - 3d - 9 = 0$.

We solve this quadratic equation for $d$: Using the quadratic formula $d = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-9)}}{2(2)}$ $d = \frac{3 \pm \sqrt{9 + 72}}{4}$ $d = \frac{3 \pm \sqrt{81}}{4}$ $d = \frac{3 \pm 9}{4}$. Two possible values for $d$ are: $d_1 = \frac{3 + 9}{4} = \frac{12}{4} = 3$. $d_2 = \frac{3 - 9}{4} = \frac{-6}{4} = -\frac{3}{2}$.

Since the roots are non-negative $0, d, 2d$, we must have $d \ge 0$. Thus, the valid value for $d$ is 3.

If $d=3$, the roots are $0, 3, 2(3) = 6$. These roots are non-negative and in AP.

Now we find the values of $a$ and $b$ using $d=3$: $a = -3d = -3(3) = -9$. $b = 2d^2 = 2(3)^2 = 2(9) = 18$.

Let's check if $a+b=9$. $a+b = -9 + 18 = 9$. This is consistent with the given condition. The polynomial is $P(x) = x^3 - 9x^2 + 18x$.

Now let's evaluate the given options based on the roots 0, 3, 6 and the values $a=-9, b=18$.

A) sum of the roots of P(x) is equal to 9. Sum of roots = $0 + 3 + 6 = 9$. This statement is true.

D) The value of (b – a) is equal to 27. $b - a = 18 - (-9) = 18 + 9 = 27$. This statement is true.