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:
sum of the roots of $P(x)$ is equal to 9.
sum of the roots of $P(x)$ is equal to 18.
the value of $(b-a)$ is equal to 9.
the value of $(b-a)$ is equal to 27.
Step-by-Step Solution
Key Concept: The key idea is to first recognize that a polynomial with no constant term ($P(x) = x^3 + ax^2 + bx$) must have $x=0$ as one of its roots. Combining this with the conditions that the roots are in arithmetic progression and are non-negative, one can deduce the specific form of the roots as $0, r, 2r$ for some $r \ge 0$, simplifying the application of Vieta's formulas.
Step 1: Let the three roots in AP be $r-d,\, r,\, r+d$ with $r-d \geq 0$.
Step 2: By Vieta's formulas for $P(x) = x^3 + ax^2 + bx$: sum of roots $= -a = (r-d)+r+(r+d) = 3r$, product of roots taken two at a time $= b = r(r-d)+r(r+d)+(r-d)(r+d) = 3r^2 - d^2$, and product of roots $= 0$ (since constant term is 0), so one root is 0.
Step 3: Since one root is 0 and roots are non-negative in AP, the roots are $0, r, 2r$ (with $d = r$). So $-a = 3r$ and $b = 0\cdot r + 0\cdot 2r + r\cdot 2r = 2r^2$.
Step 4: Sum of coefficients of $P(x)$ means $P(1) = 1 + a + b + 0 = 10$, so $1 + a + b = 10$, giving $a + b = 9$.
Step 5: Substituting $a = -3r$ and $b = 2r^2$: $-3r + 2r^2 = 9 \Rightarrow 2r^2 - 3r - 9 = 0 \Rightarrow (2r+3)(r-3) = 0$. Since $r \geq 0$, $r = 3$.
Step 6: So roots are $0, 3, 6$. Sum of roots $= 9$. Thus option (a) is correct.
Step 7: Compute $a = -3(3) = -9$ and $b = 2(9) = 18$. Then $b - a = 18 - (-9) = 27$. Thus option (d) is correct.
Correct Answer: 1, 4