Question

Question from chapter quadratic polynomial.

If sum and product of zeros of a quadratic polynomial are 1 and -30 respectively.
Then Find that polynomial.

Don't post useless things as in my previous question ​

Answer 1

《¤¤¤¤¤¤¤¤¤¤¤¤¤¤》

▪Given :-

For a Quadratic Polynomial

   

Sum of Zeros = 1

Product of Zeros = -30

___________________________

▪To Find :-

The Quadratic Polynomial.

___________________________

▪Key Point :-

If sum and product of zeros of any quadratic polynomial are s and p respectively,

Then,

The quadratic polynomial is given by :-

$\bf  {x}^{2}  - s \: x + p$

___________________________

▪Solution :-

Here,

Sum = s = 1

and

Product = p = -30

So,

Required Polynomial should be

$\bf{x}^{2}  - 1.x +  (-30)$

i.e.

$\bf  {x}^{2} -x -30$

___________________________

▪Verification :-

$\sf {x}^{2} - x - 30 \\ \\ \sf {x}^{2} - 6x + 5x - 30 \\ \\ \sf x(x - 6) + 5(x - 6) \\ \\ \sf (x - 6)(x + 5)$

So,

Zeros are 6 and -5

Sum = 6 + (-5) = 1 {VERIFIED}

Product = 6 × (-5) = -30 {VERIFIED}

___________________________

So, Required Polynomial is

$\red{ \Large\bf  {x}^{2} -x -30}$

$\Large \color{Purple}\mathfrak{  \text{W}hich \:   \: is  \:  \: the  \:  \: required }\\ \huge \color{navy} \mathfrak{ \text{ A}nswer.}$

Answer 2

Given :-

Sum of zeroes = 1

Product of zeroes = -30

To Find :-

Quadratic polynomial

Solution :-

We know that

Standard form of quadratic polynomial = x² - (α + β)x + αβ

Putting α + β = 1 and αβ = -30

x² - (1)x + (-30)

x² - 1x + (-30)

x² - 1x - 30

x² - x - 30