Question

If a and b are the roots of x² − 3x + p = 0 and c, d are the roots x² − 12x + q = 0, where a, b, c, d form a G.P. Prove that (q + p) : (q − p) = 17 : 15.

Answer 1

We have,
a +b = 3, ab = p, c + d =12 and cd = q
a, b, c and d form a G.P.
∴ First term = a,  b = ar, c = ar² and d = ar³
Then, we have
a + b = 3  and c + d = 12

\[\Rightarrow a + ar = 3 \]

\[ \Rightarrow a( 1 + r ) = 3 . . . \left( i \right)\]

\[\text { Similarly, } a r^2 (1 + r) = 12 . . . \left( ii \right)\]

\[ \Rightarrow \frac{a r^2 \left( 1 + r \right)}{a\left( 1 + r \right)} = \frac{12}{3}\]

\[ \Rightarrow r^2 = 4 \]

\[ \Rightarrow r = 2\]

\[ \therefore a \left( 1 + r \right) = 3 \]

\[ \Rightarrow a = 1\]

\[\text { Now }, p = ab \]

\[ \Rightarrow p = a \times ar = 2\]

\[\text { And, } q = cd \]

\[ \Rightarrow q = a r^2 \times a r^3 = 2^5 = 32\]

\[ \therefore \frac{q + p}{q - p} = \frac{32 + 2}{32 - 2} = \frac{34}{30} = \frac{17}{15}\]