A-LEVEL•Mathematics•Quadratics•Question 10

Solution

🟡 Exam-style · [4 marks] Show that the equation $(k+1)x^2 + 2kx + (k - 1) = 0$, where $k \ne -1$, has two distinct real roots for every value of $k$.

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Here $a = k + 1$ , $b = 2k$ , $c = k - 1$ , and $a \ne 0$ because $k \ne -1$ .

\begin{aligned} b^2 - 4ac &= (2k)^2 - 4(k + 1)(k - 1) \\ &= 4k^2 - 4(k^2 - 1) \\ &= 4k^2 - 4k^2 + 4 \\ &= 4 \end{aligned}

Since $b^2 - 4ac = 4 > 0$ for every $k$ , the equation always has two distinct real roots.

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