Question

2. Find a quadratic polynomial each with the given numbers as the sum and product of its
zeroes respectively
1. 1/4,-1​

Answer 1

Solution :

$\bf{\red{\underline{\bf{Given\::}}}}}$

The given number as the sum and product of it's zeroes respectively 1/4, -1.

$\bf{\red{\underline{\bf{To\:find\::}}}}}$

A quadratic polynomial.

$\bf{\red{\underline{\bf{Explanation\::}}}}}$

As we suppose polynomial be ax² + bx + c = 0

$\bf{\green{\underline{\underline{\bf{Sum\:of\:the\:zeroes\::}}}}}$

$\mapsto\sf{\alpha +\beta =\dfrac{-b}{a} =\dfrac{Coefficient\:of\:(x)^{2} }{Coefficient\:of\:(x)} }\\\\\\\mapsto\sf{\dfrac{1}{4} =\dfrac{-b}{a} }\\\\\\\mapsto\sf{1*a=-4b}\\\\\\\mapsto\sf{1=-4b}\\\\\\\mapsto\sf{\pink{b=-\dfrac{1}{4} }}$

$\bf{\green{\underline{\underline{\bf{Product\:of\:the\:zeroes\::}}}}}$

$\mapsto\sf{\alpha \times \beta =\dfrac{c}{a} =\dfrac{Constant\:term }{Coefficient\:of\:(x)} }\\\\\\\mapsto\sf{-1 =\dfrac{c}{a} }\\\\\\\mapsto\sf{-1*a=c}\\\\\\\mapsto\sf{-1=c}\\\\\\\mapsto\sf{\pink{c=-1 }}$

We have;

• a = 1
• b = -1/4
• c = -1

We know that formula of the quadratic equation :

$\leadsto\sf{x^{2} -(sum\:of\:zeroes)+(product\:of\:zeroes)}\\\\\leadsto\sf{x^{2} -\dfrac{1}{4} x+(-1)}\\\\\leadsto\sf{\pink{4x^{2} -x-4}}$

Answer 2

Step-by-step explanation:

Given -

Sum of zeroes = 1/4

Product of zeroes = -1

To Find -

A quadratic polynomial

As we know that :-

For polynomial :-

x² - (sum of zeroes)x + (product of zeroes)

» x² - x/4 + (-1)

» x² - x/4 - 1

» 4x² - x - 4/4 = 0

» 4x² - x - 4 = 0

Hence,

The Quadratic polynomial is 4x² - x - 4 = 0