Question

if one zero of the polynomial
3x²-8x+2k+1 is 7 times
the other then find the value
of k​

Answer 1

Let the zeroes of given polynomial be α and β

Comparing the given polynomial with ax^2+bx+c= 0

a=3

b=-8

c= 2k+1

Sum of zeroes= α+β= -b/a=8/3

Product of zeroes =αβ=c/a=(2k+1)/3

According to the question

α=7β

7β+β=8/3

8β=8/3

β=1/3

now

αβ=c/a=(2k+1)/3

7β×β=(2k+1)/3

7 (1/3)×1/3=(2k+1)/3

7/3=2k+1

2k=7/3-1

2k=4/3

k=2/3