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Question

If one zero of the polynomial 3x2 - 8x - (2k + 1) is seven times the other, find the zeroes and the value of k.


Solution

Let,
f(x) = 3x^2 - 8x - (2k + 1) and α and β be its zeroes

Here,
a = 3 
b = -8
c = -(2k + 1) 

Given,

( 1 zero of polynomial is 7 times other, 
 

α = 7β    -(i)

Sum of roots,

α + β = -b/a

7β + β = -(-8)/3    [Using (i)]

8β = 8/3

β = 1/3

Putting β = 1/3 in (i),
we have

α = 7 x 1/3 =  7/3

So,
the zeroes are:- 
α = 7/3 and β = 1/3

Now,

Product of roots = αβ =  1/3 x 7/3 = 7/9

αβ = c/a

c/a = 7/9

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

2k + 1 = -7/3

2k = -7/3-1 = -10/3

k =-10/6 =  -5/3

So,
value of k = -5/3

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