Question

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

Answer 1

Let the zero of polynomial be α,7α

sum of roots is given by

a

−b

=

3

8

8α=

3

8

α=

3

1

product of roots $

a

c

=2k+1

7α

2

=2k+1

9

7

=2k+1

2k=

9

−2

k=

9

−1

Answer 2

Answer:

Two zeros are :-

$\rm \: \dfrac{-1}{3} \: and \: \dfrac{-7}{3}$

And the value of k is :-

$\rm = \dfrac{2}{3}$

Explanation:

Given polynomial,

$\rm \implies \: {3x}^{2} + 8x + (2k + 1)$

Whose one of the zeros is 7 times the other.

If we assume one zero as α.

Then, other zero will be 7α.

On comparing the polynomial with the general form of a quadratic equation ax² + bx + c

We get,

• a = 3
• b = 8
• c = (2k + 1)

We know,

$\rm \implies \: sum \: of \: zeros = \dfrac{-b}{a}$

$\rm \implies \: \alpha + 7 \alpha = \dfrac{-8}{3}$

$\rm \implies \: 8 \alpha = \dfrac{-8}{3}$

$\rm \implies \: \alpha = \dfrac{-8}{3 \times 8}$

$\rm \implies \: \alpha = \dfrac{-1}{3}$

And also,

$\rm \implies \: product \: of \: zeros = \dfrac{c}{a}$

$\rm \implies \: \alpha \times 7\alpha = \dfrac{(2k + 1)}{3}$

$\rm \implies \: 7{ \alpha }^{2} = \dfrac{(2k + 1)}{3}$

$\rm \implies \: 7{ (\frac{-1}{3} )}^{2} = \dfrac{(2k + 1)}{3}$

$\rm \implies \: 7{ (\frac{1}{9} )} = \dfrac{(2k + 1)}{3}$

$\rm \implies \: \dfrac{7}{9} = \dfrac{(2k + 1)}{3}$

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

$\rm \implies \:21 = 18k + 9$

$\rm \implies \:21 - 9= 18k$

$\rm \implies \:12= 18k$

$\rm \implies \: \dfrac{12}{18} = k$

$\rm \implies \: \dfrac{2}{3} = k$

$\bf \therefore k = \dfrac{2}{3}$

Hence,

The zeros are :-

$\bullet \: \alpha = \dfrac{-1}{3}$

And,

$\bullet \: 7\alpha = 7 \bigg( \dfrac{-1 }{3} \bigg) = \bf \dfrac{-7}{3}$