Question

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

Answer 1

let one zero is a therefore other zero is 7a.

a+7a = 8/3

8a = 8/3

a = 1/3

now product of zero

a×7a = 7a^2

7a^2 = 2k+1/3

putting a = 1/3

7/9 = 2k+1/3

7/3 = 2k+1

6k+3 = 7

6k = 7-3 = 4

k = 4/6

k= 2/3

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Answer 2

Given that One zero of the quadratic polynomial is 7 times the other. Let these zeroes be alpha and beta.

A/Q


$\alpha = 7 \beta$


P(x) = 3x^2 -8x +2k+1

where a =3 , b=-8 and c=2k+1

We know that

$\implies { \sf \alpha + \beta = \frac{ - b}{a} }$


$7 \beta + \beta = \frac{ - ( - 8)}{3}$


$8 \beta = \frac{8}{3}$

$\beta = \frac{1}{3}$


Now, we know that


$\alpha \beta = \frac{c}{a}$


$7 \beta \times \beta = \frac{2k + 1}{3}$


$7 \times \frac{1}{3} \times \frac{1}{3} = \frac{2k + 1}{3}$


$\frac{7}{9} = \frac{2k + 1}{3}$


$7 \times 3 = (2k + 1)9$

21=18k+9

21-9 =18 k

12 =18k

12/18= k


$\boxed{ \implies \sf \: so \: \: k = \frac{2}{3} }$