Question

if one zero of a polynomial 3x2-8x+2k+1 is seven times the other.find the value of k

Answer 1

3x²-8x+2k+1

•ax²+bx+c
3x²+(-8)x+(2k+1)
Here, a=3, b=-8 ,c=2k+1

Let α,β are the zeroes of the polynomial.
α=7β

Sum of zeroes=α+β=-b/a
β+7β=-(-8)/3
8β=8/3
β=8/3÷8
β=8/3×1/8
β=1/3

Product of zeroes =αβ=c/a
(7β)(β)=(2k+1)/3
7(1/3)(1/3)=(2k+1)/3
7/9=(2k+1)/3
7/9×3=2k+1
7/3=2k+1
2k=7/3-1
2k=(7-3)/3
2k=4/3
k=4/3÷2
k=4/3×1/2
k=2/3

Hope it helps..

Answer 2

Answer:

3x²-8x+2k+1

•ax²+bx+c

3x²+(-8)x+(2k+1)

Here, a=3, b=-8 ,c=2k+1

Let α,β are the zeroes of the polynomial.

α=7β

Sum of zeroes=α+β=-b/a

β+7β=-(-8)/3

8β=8/3

β=8/3÷8

β=8/3×1/8

β=1/3

Product of zeroes =αβ=c/a

(7β)(β)=(2k+1)/3

7(1/3)(1/3)=(2k+1)/3

7/9=(2k+1)/3

7/9×3=2k+1

7/3=2k+1

2k=7/3-1

2k=(7-3)/3

2k=4/3

k=4/3÷2

k=4/3×1/2

k=2/3