If sum of the squares of zeroes of the quadratic polynomial 6x2 + x + k is 25/36, the value of k is:
(A) 4
(B) – 4
(C) 2
(D) – 2
Answers
Answered by
1
Answer:
Let α and β be the zeroes of the polynomial
⇒α
2
+β
2
=
36
25
(given)
We have (α+β)
2
=α
2
+β
2
+2αβ .....(1)
⇒α
2
+β
2
=(α+β)
2
−2αβ
We have α=
6
−1
,αβ=
6
k
Substituting the above in (1) we get
(
6
−1
)
2
−2×
6
k
=
36
25
⇒
36
1
−
6
2k
=
36
25
⇒−
6
2k
=
36
25−1
=
36
24
=
6
4
⇒−2k=4 or k=−2
Answered by
1
Step-by-step explanation:
Let α and β be the zeroes of the polynomial
⇒α
2
+β
2
=
36
25
(given)
We have (α+β)
2
=α
2
+β
2
+2αβ .....(1)
⇒α
2
+β
2
=(α+β)
2
−2αβ
We have α=
6
−1
,αβ=
6
k
Substituting the above in (1) we get
(
6
−1
)
2
−2×
6
k
=
36
25
⇒
36
1
−
6
2k
=
36
25
⇒−
6
2k
=
36
25−1
=
36
24
=
6
4
⇒−2k=4 or k=−2
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