Question

form a quadratic polynomial whose sum and product of its zeros are 1/4 and -1 respectively

Answer 1

General form of Quadratic equation is as follows :

x² - ( Sum of roots ) x + Product of roots

Given that Sum = 1 / 4 and Product = ( - 1 )

Hence Quadratic equation would be :

x² - ( 1 / 4 ) x - 1 = 0

x² - ( x / 4 ) - 1 = 0

Taking LCM we get,

4x² - x - 4 / 4 = 0

Transposing 4 from denominator to the other side we get,

4x² - x - 4 = 0 * 4 = 0

Therefore the quadratic equation is : 4x² - x - 4 = 0.

Hope it helped !

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO DETERMINE

A quadratic polynomial the sum and product of whose zeroes are

$\displaystyle \: \: \frac{1}{4} \: \: and \: \: - 1 \: \: respectively \:$

TO FIND

The quadratic polynomial

FORMULA TO BE IMPLEMENTED

The quadratic polynomial whose zeroes are given can be written as

${x}^{2} - ( \: \: sum \: \: of \: \: the \: \: zeros)x \: + \: \: ( \: product \: \: of \: \: the \: \: zeros)$

EVALUATION

The required Quadratic polynomial is

$= {x}^{2} - ( \: \: sum \: \: of \: \: the \: \: zeros)x \: + \: \: ( \: product \: \: of \: \: the \: \: zeros)$

$\displaystyle \: \: = {x}^{2} - \frac{1}{4} x - 1$

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ADDITIONAL INFORMATION

A general equation of quadratic equation is

$a {x}^{2} + bx + c = 0$

Now one of the way to solve this equation is by SRIDHAR ACHARYYA formula

For any quadratic equation

$a {x}^{2} + bx + c = 0$

The roots are given by

$\displaystyle \: x = \frac{ - b \pm \: \sqrt{ {b}^{2} - 4ac } }{2a}$