Question

12
Find the Quadratic Polynomial, whose sum and product of the zeroes are -3 and 2
respectively.
​

Answer 1

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

FORMULA TO BE IMPLEMENTED

The quadratic polynomial whose zeroes are given can be written as

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

TO DETERMINE

The Polynomial whose sum and product of the zeroes are -3 and 2 respectively

EVALUATION

The required Quadratic polynomial is

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

$= {x}^{2} - (-3)x + 2$

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

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

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

Answer 2

Answer:

$/displaystyle\huge\red{\underline{\underline{Solution}}} </h2><h2>$

Solution

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)x

2

−(sumofthezeros)x+(productofthezeros)

TO DETERMINE

The Polynomial whose sum and product of the zeroes are -3 and 2 respectively

EVALUATION

The required Quadratic polynomial is

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

2

−(sumofthezeros)x+(productofthezeros)

= {x}^{2} - (-3)x + 2=x

2

−(−3)x+2

= {x}^{2} +3x +2=x

2

+3x+2

ADDITIONAL INFORMATION

A general equation of quadratic equation is

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

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 = 0ax

2

+bx+c=0

The roots are given by

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

2a

−b±

b

2

−4ac

Step-by-step explanation:

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