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Question
Write the number of quadratic equations, with real roots, which do not change by squaring their roots.
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Solution
Let \[\alpha \text { and } \beta\] be the real roots of the quadratic equation \[a x^2 + bx + c = 0 .\]
On squaring these roots, we get: \[\alpha = \alpha^2\] and \[\beta = \beta^2\] \[\Rightarrow \alpha (1 - \alpha) = 0\] and \[\beta\left( 1 - \beta \right) = 0\]
\[\Rightarrow \alpha = 0, \alpha = 1\] and \[\beta = 0, 1\]
Three cases arise:
\[(i) \alpha = 0, \beta = 0\]
\[(ii) \alpha = 1, \beta = 0\]
\[(iii) \alpha = 1, \beta = 1\]
\[(i) \alpha = 0, \beta = 0\]
\[ \therefore \alpha + \beta = 0 \text { and } \alpha\beta = 0\]
So, the corresponding quadratic equation is,
\[x^2 - (\alpha + \beta)x + \alpha\beta = 0\]
\[ \Rightarrow x^2 = 0\]
\[(ii) \alpha = 0, \beta = 1\]
\[\alpha + \beta = 1\]
\[\alpha\beta = 0\]
So, the corresponding quadratic equation is,
\[x^2 - (\alpha + \beta)x + \alpha\beta = 0\]
\[ \Rightarrow x^2 - x + 0 = 0\]
\[ \Rightarrow x^2 - x = 0\]
\[(iii) \alpha = 1, \beta = 1\]
\[\alpha + \beta = 2\]
\[\alpha\beta = 1\]
So, the corresponding quadratic equation is,
\[x^2 - (\alpha + \beta)x + \alpha\beta = 0\]
\[ \Rightarrow x^2 - 2x + 1 = 0\]
Hence, we can construct 3 quadratic equations.
