Question

If a and B are the zeroes of polynomial p(x) = 6x² - 17x + 12, find the quadratic polynomial with zeroes 1/aplha and 1/beta.​

Answer 1

Answer:

f(x)= 6x

2

+ x− 2

a,b are its zero's / roots.

a+ b=

6

−1

,ab=

6

−2

=

−1/3

b

a

+

a

b

=

ab

2

a 2

=

ab

(a+b)

2

−2ab

=

(−1/3)

(−1/6)

2

−2(−1/3)

=

−1/3

36

1

+

3

2

=

− (

12

1

+2) =

12

−25

Answer 2

Step-by-step explanation:

Given:

$p(x) = 6 {x }^{2} - 17x + 12$

To find: Find the quadratic polynomial with zeroes $\frac{1}{\alpha}\: and\:\frac{1}{\beta}$.

Solution:

We know that relation of zeros $\alpha\:and\:\beta$ with coefficient of polynomial is given by

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

here for the given polynomial

$\alpha + \beta = \frac{17 }{6}...eq1 \\ \\ \alpha \beta = \frac{12}{6}...eq2$

Divide eq1 by eq2

$\frac{ \alpha + \beta }{ \alpha \beta } = \frac{17}{12} \\ \\ \frac{1}{ \alpha } + \frac{1}{ \beta } = \frac{17}{12} ...eq3$

and

$\frac{1}{ \alpha \beta } = \frac{6}{12} \\ \\ \frac{1}{ \alpha \beta } = \frac{1}{2}$

Therefore,

Quadratic polynomial is

${x}^{2} - \left( \frac{1}{ \alpha } + \frac{1}{ \beta } \right)x + \frac{1}{ \alpha \beta } \\ \\ {x}^{2} - \frac{17}{12} x + \frac{1}{2} = 0 \\ \\ or \\ \\ \frac{12 {x}^{2} - 17x + 6 }{12} = 0 \\ \\ 12 {x}^{2} - 17x + 6$

Final answer:

Quadratic polynomial with zeros $\frac{1}{\alpha}\: and\:\frac{1}{\beta}$ is

$\bold{\blue{12 {x}^{2} - 17x + 6}}$

Note*:In MCQs one can swap the coefficients of x² and c,in such type of questions.

Hope it helps you.

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