Question

If one zero of polynomial 3x^2+8x+2k+1 is seven times the other. then find the zeroes and value of k.

Answer 1

Hi Mate !!

Given equation :- 3x² + 8x + 2k + 1

From given equation we conclude that the Coefficient of x² is 3 , coefficient of x is 8 and constant term is ( 2k + 1 )

It is also given that one Zero is 7 times of other
So, let the first Zeros be a
then 2nd Zeros will be 7a

• Sum of Zeros =
$\frac{ - \: \: coefficient \: of \: x}{coefficient \: \: of \: {x}^{2} }$

$7a \: + \: a = \frac{-8}{3}$

$8a \: \: = \: \frac{-8}{3}$

$a \: = \: \frac{-8}{3 \times 8}$

$a \: = \frac{-1}{3}$

So, the Zeros are :- a and 7a
Then , 1st Zeros is :-
$a = \frac{-1}{3}$

and 2nd Zeros is

$7a = 7 \times \frac{-1}{3} = \frac{-7}{3}$

Now ,

• Product of Zeros =
$\frac{constant \: term}{coefficient \: of \: {x}^{2} }$

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

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

$-7 = 2k + 1$

$-7 - 1 = 2k$

$-8 = 2k \\ -4 = k$

So, the roots of the Equation are -1/3 and -7/3 & value of k is -4 !!

Answer 2

HEYA MATE,

Your solution is in the attachment.

#KEVIN

HOPE IT HELPS YOU...