Math, asked by kritisha24, 3 days ago

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'.​

Answers

Answered by maddy20007
0

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

2

=2k+1

9

7

=2k+1

2k=

9

−2

k=

9

−1

Answered by ImperialGladiator
16

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}

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