If α, β be the zeroes of the polynomial 2x2 + 5x + k such that (α�

1.	If α, β be the zeroes of the polynomial 2x2 + 5x + k such that (α + β)2 – αβ = \(\frac{21}4\) , then k = ? (a) 3 (b) -3 (c) -2 (d) 2
Answer» (d) 2 Since α and β are the zeroes of 2x² + 5x + k, we have: α + β = \(\frac{-5}2\) and αβ = \(\frac{k}2\) Also, it is given that α² + β² + αβ = \(\frac{21}4\) ⇒ (α + β)² – αβ = \(\frac{21}4\) ⇒ \((\frac{-5}2)^2\) - \(\frac{k}2\) = \(\frac{21}4\) ⇒ \(\frac{25}4\) - \(\frac{k}2\) = \(\frac{21}4\) ⇒ \(\frac{k}2\) = \(\frac{25}4\) - \(\frac{21}4\) = \(\frac{4}4\) = 1 ⇒ k = 2

If α, β be the zeroes of the polynomial 2x2 + 5x + k such that (α + β)2 – αβ = \(\frac{21}4\) , then k = ? (a) 3 (b) -3 (c) -2 (d) 2

Answer»

(d) 2

Since α and β are the zeroes of 2x² + 5x + k,

we have:

α + β = \(\frac{-5}2\) and αβ = \(\frac{k}2\)

Also, it is given that α² + β² + αβ = \(\frac{21}4\)

⇒ (α + β)² – αβ = \(\frac{21}4\)

⇒ \((\frac{-5}2)^2\) - \(\frac{k}2\) = \(\frac{21}4\)

⇒ \(\frac{25}4\) - \(\frac{k}2\) = \(\frac{21}4\)

⇒ \(\frac{k}2\) = \(\frac{25}4\) - \(\frac{21}4\) = \(\frac{4}4\) = 1

⇒ k = 2