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Question

If one zero of 3x2-8x+2k+1 is 7 times the other, find the value of k.

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Solution

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
therefore k=2/3

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