Question

22. Find the quadratic polynomial whose zeroes are -1/3
and 2. and with method
​

Answer 1

Step-by-step explanation:

I hope this is the answer

Answer 2

Question-

Find the quadratic polynomial whose zeroes are

-1/3 and 2

Solution -

Here,

$\alpha = \frac{ - 1}{3}$

$\beta = 2$

Equation is,

$\implies(x - \alpha )(x - \beta ) = 0$

$\implies(x - \frac{ - 1}{3} )(x - 2) = 0$

$\implies(x + \frac{1}{3} )(x - 2) = 0$

$\implies \: x(x + \frac{1}{3} ) - 2 (x + \frac{1}{3} ) = 0$

$\implies \: {x}^{2} + \frac{x}{3} - 2x - \frac{2}{3} = 0$

$\implies \: {x}^{2} + ( \frac{x - 6x}{3}) - \frac{2}{3} = 0$

$\implies \: {x}^{2} + ( \frac{ - 5x}{3} ) - \frac{2}{3} = 0$

$\implies \: {x}^{2} - \frac{5x}{3} - \frac{2}{3} = 0$

$\therefore \: the \: quadratic \: equation \: whose \: zeroes \: are \: \: \frac{ - 1}{3} \\ and \: 2 \: is \: {x}^{2} - \frac{5x}{3} - \frac{2}{3} = 0$