Find the value of $m$ for which the quadratic equation $(m-1)x^2 + 2(m-1)x + 1 = 0$ has two real and equal roots.
Generated by claude-sonnet-4-6 · 2026-06-15 10:31 · grounding rag
Model Answer
For two equal real roots, discriminant $D = b^2 - 4ac = 0$.
Here, $a = (m-1)$, $b = 2(m-1)$, $c = 1$.
$$D = [2(m-1)]^2 - 4(m-1)(1) = 0$$
$$4(m-1)^2 - 4(m-1) = 0$$
$$4(m-1)[(m-1) - 1] = 0$$
$$4(m-1)(m-2) = 0$$
So, $m = 1$ or $m = 2$.
But if $m = 1$, the coefficient of $x^2$ becomes 0, so it is no longer a quadratic equation.
Therefore, $m = 2$.
Source: Chapter 4, Section 4.4 — Nature of Roots
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Explanation
- The key condition for equal roots is $D = b^2 - 4ac = 0$.
- Examiners expect you to reject $m = 1$ explicitly, because $a = m - 1 = 0$ would make it non-quadratic ($a \neq 0$ is mandatory).
- Both steps — setting up $D = 0$ and rejecting $m = 1$ — carry marks, so don't skip either.