Question

2. Each question 1 21. Find a quadratic polynomial with zeroes -2 and 1/3​

Answer 1

Given:

Zeroes are  -2 and $\frac{1}{3}$ or we can say

• α = -2

• β =  $\frac{1}{3}$

To find:

A quadratic polynomial with zeroes  -2 and $\frac{1}{3}$

Solution:

• The form of a quadratic equation is $ax^{2}$ + bx + c .

• Another form of a quadratic equation is $x^{2} - (\alpha + \beta )x + \alpha \beta = 0$

Sum of Zeroes:

= α+β

= -2 + $\frac{1}{3}$

= $\frac{-6}{3}$ +  $\frac{1}{3}$

= $\frac{-6+1}{3}$

= $\frac{-5}{3}$

Product of zeroes:

= αβ

= -2 ×  $\frac{1}{3}$

= $\frac{-2}{3}$

By substituting the values, α+β = $\frac{-5}{3}$  and αβ = $\frac{-2}{3}$

$x^{2} - (\alpha + \beta )x + \alpha \beta = 0$

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

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

Multiplying both sides of the equation by 3, we get:

$3x^{2} + 5x - 2 = 0$

Final Answer:

The required quadratic polynomial is $3x^{2} + 5x - 2$ .