Question

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

Answer 1

let zeroes be a and b
a =7b
a+b=8/3
7b+b=8/3
8b=8/3
b=1/3
b is the zero of the polynomial
so
3(1/3)^2 - 8(1/3) + 2k+1=0
3/9 - 8/3 +2k+1=0
3 - 24/9 + 2k +1 = 0
-21 +18 k +9 =0
18k=21-9
18k=12
k=12/18
k= 2/3

Answer 2

p(x) = 3x² - 8x + (2k+1)

According to question let a zero be α and the other would be 7α
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Sum of zeroes = -b/a

α + 7α = -(-8)/3

      8α = 8/3

        α = 1/3
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Product of zeroes = c/a

α × 7α = (2k+1)/3

 3(7α²) = 2k + 1 

Substituting α = 1/3

3 × 7 × (1/3)² = 2k + 1

                7/3 = 2k + 1

         3(2k+1) = 7

            6k + 3 = 7

               6k = 7 - 3

                 k = 4/6  ⇒ k = 2/3

Hence k = 2/3 
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