Question

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

• A. 3
• B. 1
• C. 4
• D. 0

Answer 1

The Correct Option is C

Given that \( (3+2\sqrt{2}) \) is a root of the equation \( ax^2 + bx + c = 0 \), its conjugate \( (3-2\sqrt{2}) \) is also a root. Thus, the equation can be expressed as: \[ ax^2 + bx + c = a(x - (3+2\sqrt{2}))(x - (3-2\sqrt{2})) \] Expanding the factored form: \[ = a((x-3)^2 - (2\sqrt{2})^2) \] \[ = a(x^2 - 6x + 1) \] Thus, comparing coefficients: - \( b = -6a \) - \( c = a \) Now, \( (4 + 2\sqrt{3}) \) is a root of the equation \( ay^2 + my + n = 0 \) with its conjugate \( (4 - 2\sqrt{3}) \) also a root. Therefore: \[ ay^2 + my + n = a(y - (4+2\sqrt{3}))(y - (4-2\sqrt{3})) \] Expanding the factored form: \[ = a((y-4)^2 - (2\sqrt{3})^2) \] \[ = a(y^2 - 8y + 4) \] Thus, comparing coefficients: - \( m = -8a \) - \( n = 4a \) Now, we find the expression \(\left(\frac{b}{m} + \frac{c-2b}{n}\right)\): \[ \frac{b}{m} = \frac{-6a}{-8a} = \frac{3}{4} \] \[ \frac{c-2b}{n} = \frac{a - 2(-6a)}{4a} = \frac{13a}{4a} = \frac{13}{4} \] Adding these: \[ \frac{3}{4} + \frac{13}{4} = \frac{16}{4} = 4 \] Conclusively, the value is \(4\).