Doubt by Saksham
Solution :
2x²+5x+k
a = 2
b = 5
c = k
α+β = -b/a
α+β = -(5)/2
α+β = -5/2 — (1)
αβ = c/a
αβ = k/2 — (2)
α²+β²+αβ = 21/4 (Given)
[(α+β)²-2αβ]+αβ = 21/4 [∵α²+β²=(α+β)²-2αβ]
(α+β)²-2αβ+αβ = 21/4
α+β = -b/a
α+β = -(5)/2
α+β = -5/2 — (1)
αβ = c/a
αβ = k/2 — (2)
α²+β²+αβ = 21/4 (Given)
[(α+β)²-2αβ]+αβ = 21/4 [∵α²+β²=(α+β)²-2αβ]
(α+β)²-2αβ+αβ = 21/4
(α+β)²-αβ = 21/4
(-5/2)²-k/2 = 21/4 [Using (1) and (2)]
25/4-k/2 = 21/4
25/4-21/4 = k/2
(25-21)/4 = k/2
4/4 = k/2
1 = k/2
1×2 = k
2 = k
k = 2
(-5/2)²-k/2 = 21/4 [Using (1) and (2)]
25/4-k/2 = 21/4
25/4-21/4 = k/2
(25-21)/4 = k/2
4/4 = k/2
1 = k/2
1×2 = k
2 = k
k = 2
Hence, the required value of k is 2.