Question

find a quadratic polynomial each with the given numbers as the sum and the product of its zeroes. (i) 2/3,-8/9​

Answer 1

Answer:

Given :- 2/3,-8/9 find a quadratic

polynomial each with the given numbers as the sum and product of its zeroes respectively?

Answer :

given that,

• sum of zeroes = (2/3)

product of zeroes = (-8/9)

SO,

➡ x² - (sum of zeroes)x+ product of zeroes = 0

→x² - (2/3)x+ (-8/9) = 0

→x² - (2x/3) - (8/9) = 0

→ (9x² - 6x - 8) /9 = 0

- 9x² - 6x - 8 = 0 (Ans.)

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

Given:

Sum of zeros : 2/3

Product of zeros: -8/9

To find: Find the quadratic polynomial.

Solution:Let a quadratic polynomial have zeros $\alpha\:and\:\beta$ then it is given by

$\bold{ {x}^{2} - ( \alpha + \beta )x + \alpha \beta}$

Here,

Sum of zeros

$\alpha + \beta = \frac{2}{3}$

Product of zeros

$\alpha \beta = \frac{ - 8}{9}$

Therefore quadratic polynomial is

${x}^{2} - \frac{2}{3} x +\left( \frac{ - 8}{9} \right) \\ \\ or \\ \\ {x}^{2} - \frac{2}{3} x - \frac{8}{9} \\ \\ or \\ \\ \frac{9{x}^{2} - 6x -8 }{9} = 0 \\ \\ 9{x}^{2} - 6x - 8= 0$

Final answer:

The polynomial is

$\boxed{\bold{\green{9 {x}^{2} - 6x - 8 = 0 }}}$

whose sum of zeros is 2/3 and product of zeros are -8/9

Hope it helps you.

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