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If \(\alpha,\beta\) are the zeroes of kx2 – 2x + 3k such that \(\alpha + \beta\) = \(\alpha\,\beta\), then find k? |
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Answer» Given that \(\alpha,\beta\) are zeros of kx2 – 2x + 3k such that \(\alpha + \beta\) = \(\alpha\,\beta\). Since, α & β are zeros of kx2 – 2x + 3k. Therefore, sum of zeros = \(\alpha + \beta\) = \(\frac{-b}{a}\) = \(\frac{-(-2)}{k}\) = \(\frac{2}{k}\). (In kx2 – 2x + 3k, a = k, b = –2, c = 3k) & product of zeros \(\alpha,\beta\) = \(\frac{c}{a} = \frac{3k}{k} = 3\). Now, given that \(\alpha + \beta\) = \(\alpha\,\beta\) ⇒ \(\frac{2}{k}\) = 3. Hence, k = \(\frac{2}{3}\) . If α,β are the zeroes of kx2 – 2x + 3k then α+β=2/k and αβ=3k/k=3 again we have α+β=αβ so 2/k =3 =>k=2/3 |
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