Question

Find the quadratic polynomial whose zeroes are -1/3 and 1/3

Answer 1

Answer:

if the zeroes are -1/3 and 1/3

$\alpha$+$\beta$= -b/a= -1/3

$\alpha \beta$=c/a=1/3

therefore,the required polynomial is,

3$x^{2}$+x+1

Step-by-step explanation:

hope it helps

Answer 2

Answer: All the quadratic polynomials of form   $a(x^{2} - \frac{1}{9})$ with the condition a ≠ 0, will have -1/3 and 1/3 as roots

Step-by-step explanation:

Let the quadratic polynomial be ax² + bx + c ( a ≠ 0 )

Roots of the polynomial  ⇒ -1/3 , 1/3

Sum of roots =  $\frac{-b}{a}$

$\frac{-1}{3} + \frac{1}{3} = \frac{-b}{a}$

$0 = \frac{-b}{a}$

b = 0

Product of roots =  $\frac{c}{a}$

$\frac{-1}{3} * \frac{1}{3} = \frac{c}{a}$

$\frac{-1}{9} = \frac{c}{a}$

$c = \frac{-a}{9}$

 

Quadratic polynomial ⇒ ax² + bx + c

Quadratic polynomial ⇒  $ax^{2} - \frac{a}{9}$

Quadratic polynomial ⇒  $a(x^{2} - \frac{1}{9})$

All the polynomials of form   $a(x^{2} - \frac{1}{9})$ will have -1/3 and 1/3 as roots.

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