55 views

If $(3+2 \sqrt{2})$ is a root of the equation $a x^{2}+b x+c-0$, and $(4+2 \sqrt{3})$ is a root of the equation $a y^{2}+m y+n – 0$, where $a, b, c, m$ and $n$ are integers, then the value of $\left(\frac{b}{m}+\frac{c-2 b}{n}\right)$ is

1. $0$
2. $3$
3. $4$
4. $1$

The roots of ax$^{2}$+bx+c=0 are (-b+$\sqrt{b^{2} -4ac}$)/2a and (-b-$\sqrt{b^{2} -4ac}$)/2a. And one of the root is (3+2$\sqrt{2}$). The value (3+2$\sqrt{2}$) is equal to (-b+$\sqrt{b^{2} -4ac}$)/2a because sign of irrational part of both are same. ($\sqrt{2}$ is irrational)

Compare rational part and irrational part separately,

(-b+$\sqrt{b^{2} -4ac}$)/2a = (3+2$\sqrt{2}$)

=> (-b+$\sqrt{b^{2} -4ac}$) = (6a+$\sqrt{32a^{2}}$)  => b = -6a  &  c = a (b$^{2}$-4ac = 32a$^{2}$  => 36a$^{2}$ – 4ac = 32a$^{2}$  => c = a).

Similarly we have to do for 2nd euqation :-

The roots of ax$^{2}$+mx+n=0 are (-m+$\sqrt{m^{2} -4an}$)/2a and (-m-$\sqrt{m^{2} -4an}$)/2a. And one of the root is (4+2$\sqrt{3}$). The value (4+2$\sqrt{3}$) is equal to (-m+$\sqrt{m^{2} -4an}$)/2a because sign of irrational part of both are same. ($\sqrt{3}$ is irrational)

Compare rational part and irrational part separately,

(-m+$\sqrt{m^{2} -4an}$)/2a = (4+2$\sqrt{3}$)

=> (-m+$\sqrt{m^{2} -4an}$) = (8a+$\sqrt{48a^{2}}$)  => m = -8a  &  n = 4a (m$^{2}$-4an = 48a$^{2}$  => 64a$^{2}$ – 4an = 48a$^{2}$  => n = 4a).

Therefore the value of (b/m)+(c-2b)/n = (-6a/-8a)+(a+12a)/4a = 4.

by
154 points

1 vote
1
1 vote