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Let $x_1$ and $x_2$ be the roots of the quadratic equation $x^2-3x+a=0$, and $x_3$ and $x_4$ be the roots of the quadratic equation $x^2-12x+b=0$. If $x_1, x_2, x_3$ and $x_4 \: (0 < x_1 < x_2 < x_3 < x_4)$ are in $G.P.,$ then $ab$ equals

1. $64$
2. $5184$
3. $-64$
4. $-5184$

edited | 83 views

Answer: $\mathbf A$

Explanation:

For equation $1$:

Let the $\mathbf {GP}$ be:

$\mathrm {A, AR, AR^2, AR^3},\dots$ (I choose capital letters because a and b already used in the question)
Sum of roots$= x_1 + x_2 = \frac{-b}{a} = 3 \Rightarrow a + ar = 3 \Rightarrow a(1+r) = 3$ $\cdots \tag{1}$

Product of roots $= x_1.x_2 = \frac{c}{a} = a \cdots \tag{2}$

Similarly, for equation $2$:

Sum of roots $= x_3 + x_4 =12 \Rightarrow ar^2+ar^3 = ar^2(1+r) = 12 \cdots\tag{3}$

Product of roots $= x_3.x_4 = b \cdots \tag{4}$

Now, since all the roots are in $GP$

Dividing equation $1$ by $2$ we get:

$\frac{1}{r^2} = \frac{1}{4} \Rightarrow r = \pm2 \Rightarrow r = 2$

Now,

Substitute the value of r in equation 1,we get:

$A(1+R) = 3 \Rightarrow A(1+2) = 3 \Rightarrow A = 1$

So, $A = 1, R = 2$

From equation $2$

$x_1.x_2 = a \Rightarrow A.AR = a$

Now,

$x_1.x_2 = a \Rightarrow x_1.x_2 = a.ar = 1.1.2 = 2$

similarly,

$x_3.x_4 = b \Rightarrow ar^2.ar^3 = a^2r^5 = 1^2.2^5 = 32$

So, $ab = 2.32 = 64$

Hence, option $\mathbf A$ is the right answer.

by Boss (19.4k points)
edited by