If the zeros of x 2 kx + 6 are in the ratio 3:2 , find k.
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Answered by
41
Let α,β are the zeros of the polynomial x²+kx+6. Then,
α+β=-k/1=-k and αβ=6/1=6
By the given condition,
α:β=3:2
or, α=3β/2
∴, 3β/2+β=-k
or, (3β+2β)/2=-k
or, 5β=-2k
or, β=-2k/5
3β/2×β=6
or, 3β²=12
or, β²=4
or, (-2k/5)²=4
or, 4k²/25=4
or, 4k²=100
or, k²=25
or, k=+-5
[There is no sign mentioned between x² and kx therefore I assume it as positive sign.]
α+β=-k/1=-k and αβ=6/1=6
By the given condition,
α:β=3:2
or, α=3β/2
∴, 3β/2+β=-k
or, (3β+2β)/2=-k
or, 5β=-2k
or, β=-2k/5
3β/2×β=6
or, 3β²=12
or, β²=4
or, (-2k/5)²=4
or, 4k²/25=4
or, 4k²=100
or, k²=25
or, k=+-5
[There is no sign mentioned between x² and kx therefore I assume it as positive sign.]
Answered by
1
Answer:
=> k = 5 or k = -5
Step-by-step explanation:
Let roots are α and β
αβ=32
=> α=32β.....(1)
Then
Product of roots = c/a
=> αβ=6
=> 32ββ=6
=> β2=4
=> β=±2
Putting in (1) we get
α=±3
Now sum of roots = −b/a
=> 3 + 2 = k or -3 + (-2) = k
=> k = 5 or k = -5
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