Question

if one zero of polynomial 3x^2=8x+2k+1 is 7 times the other find the zeroes and value of k​

Answer 1

Solution:

Let m, 7m are two roots of

given Quadratic equation

3x²=8x+2k+1

=> 3x²-8x-(2k+1)=0

Compare this with ax²+bx+c=0

we get

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

i ) sum of the roots = -b/a

=> m+7m = -(-8)/3

=> 8m = 8/3

=> m = (8/3) × 1/8

=> m = 1/3 ---(1)

ii ) product of the roots = c/a

=> m×7m = -(2k+1)/3

=> 7m² = -(2k+1)/3

=> 7(1/3)² = -(2k+1)/3

=> 7/9 × 3 = -(2k+1)

=> 7/3 = -2k-1

=> 7/3+1= -2k

=> (7+3)/3 = -2k

=>10/3 = -2k

=> 10/(-6) = k

Therefore,

k = (-10/6) = -5/3

•••••

Answer 2

Hey mate.

Here,

3x²=8x+2k+1=0

where,

3x²-8x+2k+1=0

Where as

values of a,b and c are,

3,-8 and 2k+1

Let be the condition for it

let be alpha and ß as zeros of the given polynomial

One condition for this ,

alpha =7ß

alpha + ß =-b/a

so,

7ß+ß= [ -(-8)/3 ]

8ß= 8/3

ß=8/3 ×8

ß=1/3

alpha ×ß = c/a

7ß×ß= 2k+1/3

7(1/3)×1/3 =2k+1

7/9=2k+1

21/9 = 2k+1

21/9-1 = 2k

21-9/9= 2k

12/9= 2 k

4/3= 2k

4/6=k

2/3=k.

Hope it will help you

✌️sai