Question:
Let a, b \(\in\) R, a \(\ne\) 0 be such that the equation, \(ax^2 - 2bx + 5 = 0\) has a repeated root \(\alpha\), which is also a root of the equation, \(x^2 - 2bx - 10 = 0\). If \(\beta\) is the other root of this equation, then \(\alpha^2 + \beta^2\) is equal to :
Let a, b \(\in\) R, a \(\ne\) 0 be such that the equation, \(ax^2 - 2bx + 5 = 0\) has a repeated root \(\alpha\), which is also a root of the equation, \(x^2 - 2bx - 10 = 0\). If \(\beta\) is the other root of this equation, then \(\alpha^2 + \beta^2\) is equal to :
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A repeated root means the discriminant is zero, which gives \(b^2 = 5a\) and \(\alpha = b/a\). Substitute \(\alpha\) into the second equation to find a, then use sum and product of roots.
Updated On: Nov 5, 2026
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The Correct Option is D
Solution and Explanation
Equation 1 has equal roots, so \(D = 4b^2 - 20a = 0\), i.e. \(b^2 = 5a\), and the repeated root is \(\alpha = \frac{2b}{2a} = \frac{b}{a}\).
\(\alpha\) satisfies \(x^2 - 2bx - 10 = 0\): \(\frac{b^2}{a^2} - \frac{2b^2}{a} - 10 = 0\).
Using \(b^2 = 5a\): \(\frac{5}{a} - 10 - 10 = 0\), so \(a = \frac{1}{4}\) and \(b^2 = \frac{5}{4}\).
For the second equation, \(\alpha + \beta = 2b\) and \(\alpha\beta = -10\).
\(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 4b^2 + 20 = 5 + 20 = 25\).
Answer: (4) 25
\(\alpha\) satisfies \(x^2 - 2bx - 10 = 0\): \(\frac{b^2}{a^2} - \frac{2b^2}{a} - 10 = 0\).
Using \(b^2 = 5a\): \(\frac{5}{a} - 10 - 10 = 0\), so \(a = \frac{1}{4}\) and \(b^2 = \frac{5}{4}\).
For the second equation, \(\alpha + \beta = 2b\) and \(\alpha\beta = -10\).
\(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 4b^2 + 20 = 5 + 20 = 25\).
Answer: (4) 25
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