For what value of k will the roots of the equation kx2 – 5x + 6 =

1.	For what value of k will the roots of the equation kx2 – 5x + 6 = 0 be in the ratio 2 : 3 ?(a) 0 (b) 1 (c) –1 (d) 2
Answer» (b) 1 Let the roots of the equation kx² – 5x + 6 = 0 be α and β. Then, α + β = \(\frac{5}{k}\) ...(i) αβ = \(\frac{6}{k}\) ...(ii) Given \(\frac{α}{β}\) = \(\frac{2}{3}\) ⇒ α = \(\frac{2}{3}\)β ∴ From (i) and (ii), \(\frac{2}{3}\)β + β = \(\frac{5}{k}\) and \(\frac{2}{3}\)β² = \(\frac{6}{k}\) ⇒ \(\frac{5}{3}\)β = \(\frac{5}{k}\) and β² = \(\frac{9}{k}\) ⇒ β = \(\frac{3}{k}\) and β² = \(\frac{9}{k}\) ⇒ \(\frac{9}{k^2}\) = \(\frac{9}{k}\) ⇒ 9k² - 9k = 0 k(k - 1) = 0 ⇒ k = 0 or 1 But k = 0 does not satisfy the condition, so k = 1.

For what value of k will the roots of the equation kx2 – 5x + 6 = 0 be in the ratio 2 : 3 ?(a) 0 (b) 1 (c) –1 (d) 2

Answer»

(b) 1

Let the roots of the equation kx² – 5x + 6 = 0 be α and β.

Then, α + β = \(\frac{5}{k}\) ...(i)

αβ = \(\frac{6}{k}\) ...(ii)

Given \(\frac{α}{β}\) = \(\frac{2}{3}\) ⇒ α = \(\frac{2}{3}\)β

∴ From (i) and (ii),

\(\frac{2}{3}\)β + β = \(\frac{5}{k}\) and \(\frac{2}{3}\)β² = \(\frac{6}{k}\)

⇒ \(\frac{5}{3}\)β = \(\frac{5}{k}\) and β² = \(\frac{9}{k}\) ⇒ β = \(\frac{3}{k}\) and β² = \(\frac{9}{k}\)

⇒ \(\frac{9}{k^2}\) = \(\frac{9}{k}\) ⇒ 9k² - 9k = 0 k(k - 1) = 0 ⇒ k = 0 or 1

But k = 0 does not satisfy the condition, so k = 1.