Question

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

Answer 1

3x²-8x+2k+1 a=3, b=-8 ,c=2k+1
suppose the zeroes of the equation are α and β . by the given condition α=7β
as we know , α+β=-b/a
7β+β=-(-8)/3
8β=8/3
β=8/3×1/8=1/3
αβ=c/a
7β×β=2k+1/3
7×1/3×1/3=2k+1/3
7/9=2k+1/3
21/9=2k+1
21/9-1=2k
12/9=2k
4/3=2k
4/6=k
2/3=k

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Answer 2

HEYA!!
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Let the zeroes be a and 7 a respectively .

Sum of zeroes = 7a + a = 8a .

Sum of zeroes = -b/a

8a = 8/3

24a = 8

a = 1/3

Thus , the zeroes are 1/3 and 7/3

Putting X = 1/3 in the given equation ,,

$3 \times \frac{1}{9} - 8 \times \frac{1}{3} + 2k + 1 \\ \\ \frac{1}{3} - \frac{8}{3} + 1 + 2k = 0 \\ \\ \frac{1 - 8 + 3 + 6k}{3} = 0 \\ \\ \frac{6k - 4}{3} = 0 \\ \\ 6k - 4 = 0 \\ \\ 6k = 4 \\ \\ k = \frac{2}{3}$
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