Bunuel wrote:
The quadratic equations \(x^2 − 6x + a = 0\) and \(x^2 − cx + 6 = 0\) have one root in common. If the other roots of the first and second equations are integers in the ratio 4:3, then the common root is
A. 1
B. 2
C. 3
D. 4
E. 5
Good Question!
Let the common root = \(m\) & the other roots be \(4k\) & \(3k\) respectively
--> The roots of 1st equation are (\(m, 4k\)) and the roots of the 2nd equation are (\(m, 3k\))
Formula: For the quadratic equation \(ax^2 + bx + c = 0\), Sum of the roots = \(\frac{-b}{a}\) & Product of the roots = \(\frac{c}{a}\)For \(x^2 − 6x + a = 0\),
Sum of the roots = \(m + 4k\) = \(\frac{-(-6)}{1} = 6\)
--> \(m + 4k = 6\) ........ (1)
and for \(x^2 − cx + 6 = 0\)
Product of the roots = \(m*3k\) = \(\frac{6}{1}\)
--> \(m*k = 2\)
--> \(k = \frac{2}{m}\) ........ (2)
Substituting (2) in (1),
--> \(m + 4(2/m) = 6\)
--> \(m^2 + 8 = 6m\)
--> \(m^2 - 6m + 8 = 0\)
--> \((m - 2)(m - 4) = 0\)
--> \(m = 2\) or \(4\)
Given that, all roots should be integers
Case 1: If \(m = 2, k = \frac{2}{m} = 1\), So, other roots are \(4k\) & \(3k\) = \(4\) & \(3\) respectively -->
PossibleCase 2: If \(m = 4, k = \frac{2}{m} = \frac{1}{2}\), So, other roots are \(4k\) & \(3k\) = \(2\) & \(3/2\) respectively -->
Not PossibleSo, \(m = 2\) ONLY
Option B