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If \( x^{2}-a x+a-1=0 \) and \( x^{2}-4 x+a=0 \) have a common root then find the sum of all possible value(s) of a |
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Answer» Both quadratic equations \(x^2 - ax + a - 1 = 0\) and \(x^2 - 4x + a = 0\) have a common root. Let \(\alpha\) be common root \(\therefore \alpha^2 - 4\alpha + a = 0\) and \(\alpha^2 - a\alpha + a - 1 = 0\) \(\Rightarrow \alpha^2 - 4\alpha + a\) \(= \alpha^2 - a\alpha + a - 1\) \(\Rightarrow a\alpha\, - 4\alpha = -1\) \(\Rightarrow \alpha = \frac{-1}{a - 4}.\) i.e., both equations have common root \(x = \frac{-1}{a - 4}.\) \(\therefore\) \(\left(\frac{-1}{a - 4}\right)^2 - a\left(\frac{-1}{a - 4}\right)\) \(+\, a - 1 = 0\) \(\Rightarrow \frac{1}{(a - 4)^2} + \frac{a}{a - 4}\) \(+\, a - 1 = 0\) \(\Rightarrow (1 + a)(a - 4)\) \(+\, (a - 1)(a + 4)^2 = 0\) \(\Rightarrow\) (a - 1) \((a^2 - 8a + 16)\) \(+\, a^2 - 4a + 1 = 0\) \(\Rightarrow\) \(a^3 - 9a^2 + 24a - 16\) \(+\, a^2 - 4a + 1 = 0\) \(\Rightarrow\) \(a^3 - 8a^2 + 20a - 15 = 0\) sum of all possible values of a \(= \frac{-(-8)}{1} = 8.\) |
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