Question

find quadratic polynomial whose zeroes are 2/3 and -8/9​

Answer 1

Step by step Explanation:

Given: Zeros of polynomial 2/3 and -8/9

To find: Quadratic polynomial

Solution:

If $\alpha\:\:and\:\:\beta$ are zeros of polynomial than quadratic polynomial is

${x}^{2} - ( \alpha + \beta )x + \alpha \beta ...eq1$

Thus,

Find the sum and multiplication of zeros,

Let

$\alpha = \frac{2}{3} \\ \\ \beta = - \frac{8}{9} \\ \\ \alpha + \beta = \frac{2}{3} + \frac{ - 8}{9} \\ \\ \alpha + \beta = \frac{6 - 8}{9} \\ \\ \alpha + \beta = \frac{ - 2}{9}...eq2 \\ \\ \alpha \beta = \frac{2}{3} \times ( \frac{ - 8}{9} ) \\ \\ \alpha \beta = \frac{ - 16}{27}...eq3$

Put the values from eq2 and eq3 in eq1

${x}^{2} - ( - \frac{2}{9} )x + ( - \frac{16}{27} ) \\ \\ {x}^{2} + \frac{2}{9} x - \frac{16}{27}$

Therefore,

Quadratic polynomial is

${x}^{2} + \frac{2}{9} x - \frac{16}{27}$

or

$\frac{27 {x}^{2} + 6x - 16}{27} = 0 \\ \\ 27 {x}^{2} + 6x - 16 = 0$

Final answer:

$\bold{27 {x }^{2} + 6x - 16}$

Hope it helps you.

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