If \((3+2\sqrt2)\) is a root of the equation ax2 + bx + c = 0 and \((4+2\sqrt3)\) is a root of the equation ay2 + my + n = 0,where a,b,c,m and n are integers, then the value of \((\frac{b}{m}+\frac{c−2b}{n})\) is
Given that \( (3+2\sqrt{2}) \) is a root of \( ax^2 + bx + c = 0 \), its conjugate \( (3-2\sqrt{2}) \) is also a root. Thus, the equation is \( ax^2 + bx + c = a(x - (3+2\sqrt{2}))(x - (3-2\sqrt{2})) \). Expanding yields \( a((x-3)^2 - (2\sqrt{2})^2) = a(x^2 - 6x + 1) \). Comparing coefficients: \( b = -6a \) and \( c = a \).
Given that \( (4 + 2\sqrt{3}) \) is a root of \( ay^2 + my + n = 0 \), its conjugate \( (4 - 2\sqrt{3}) \) is also a root. Thus, the equation is \( ay^2 + my + n = a(y - (4+2\sqrt{3}))(y - (4-2\sqrt{3})) \). Expanding yields \( a((y-4)^2 - (2\sqrt{3})^2) = a(y^2 - 8y + 4) \). Comparing coefficients: \( m = -8a \) and \( n = 4a \).
We calculate \( \left(\frac{b}{m} + \frac{c-2b}{n}\right) \).
\( \frac{b}{m} = \frac{-6a}{-8a} = \frac{3}{4} \)
\( \frac{c-2b}{n} = \frac{a - 2(-6a)}{4a} = \frac{13a}{4a} = \frac{13}{4} \)
Adding these: \( \frac{3}{4} + \frac{13}{4} = \frac{16}{4} = 4 \).
The value is \(4\).