Question

If equation $16x^4 - mx^3 + (2m + 17)x^2 - mx + 16 = 0$ has four distinct roots forming a geometric progression, and one of the root is $\frac{1}{2}$, then :

• A. common ratio of G.P. is 4
• B. m = 170
• C. sum of the roots = $\frac{85}{8}$
• D. Product of root = 3

Answer 1

Answer: (B), (C)

m = 170, sum of the roots = $\frac{85}{8}$

Answer 2

The given equation is a reciprocal equation of type I. If $\alpha$ is a root, then $1/\alpha$ is also a root. Since $\frac{1}{2}$ is a root, $2$ must also be a root.

Let the four distinct roots forming a geometric progression be $a/r^3, a/r, ar, ar^3$. The product of the roots is $a^4$. From the equation, the product of the roots is $16/16 = 1$. So, $a^4 = 1 \implies a = \pm 1$.

Using $a=1$, the roots are $1/r^3, 1/r, r, r^3$. Since $1/2$ and $2$ are roots, the only way for them to fit this pattern (with distinct roots and product 1) is if $1/r = 1/2$ (or $r=2$). This implies $r=2$.

The roots are then $1/2^3, 1/2, 2, 2^3$, which are $1/8, 1/2, 2, 8$.

• Common ratio of G.P. is 4: False, it is 2.
• Sum of roots $= 1/8 + 1/2 + 2 + 8 = (1+4+16+64)/8 = 85/8$.
• From the equation, sum of roots $= -(-m/16) = m/16$.
• Equating sums: $m/16 = 85/8 \implies m = 170$. So, "m = 170" is true and "sum of the roots = 85/8" is true.
• Product of root = 3: False, product of roots is 1.

Therefore, m = 170 and sum of the roots = $\frac{85}{8}$ are the correct options.