Question

2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeros respectively

Answer 1

Solution:-

given by:-
(i) roots (2 , 1/5)
sum of roots = root 2
product of rooots = 1/5

Quadratic polynomial :-

x² - [ sum of roots]x + [ product of roots ] =0
x² - [ root2 ]x + [1/5 ]

(5x² - 5root2 x + 1 =0)ans

(ii) roots (1/3 , 1)
sum of roots = 1/3
product of roots = 1

Quadratic polynomial :-

x² - [ sum of roots ]x + [ product of roots ] =0
x² - [ 1/3 ]x + [1 ] =0

(3x² - x + 3 =0) ans

(iii) roots (1/2 , -4)
sum of roots = 1/2
product of roots = -4

Quadratic polynomial :-

x² - [ sum of roots ]x + [ product of roots ] =0
x² - [ 1/2 ]x + [-4 ] =0

(2x² - x -8 =0) ans

(iv) roots (1/3 , -1/3)
sum of roots = 1/3
product of roots = -1/3

Quadratic polynomial :-

x² - [ sum of roota ]x + [ product of roots ] =0
x² - [ 1/3 ]x + [-1/3 ]=0

(3x² - x - 1 =0) ans

(v) roots (root 3 , 2)
sum of roots = root 3
product of roots = 2

Quadratic polynomial :-

x² - [ sum of roots ]x + [ product of roots ] =0
x² - [ root3 ]x + [2 ] =0

(x² - root3 x + 2 =0) ans

Answer 2

$\large{\mathbf{HELLO \: \: DEAR,}}$


$\mathit{FORMULA \: \: FOR \: \: QUADRATIC \: \: EQUATIONS \: \: IS \: \: } \\ \mathit{X^2 - (\alpha + \beta)x + \alpha*\beta=0}$

$\it{\boxed{4} . \: \: \alpha + \beta = \sqrt{2}} \\ \\ \it{\alpha * \beta = \frac{1}{5}}$

$\mathit{X^2 - \sqrt{2}x + \frac{1}{5}=0}\\ \\ \mathit{5x^2 - 5\sqrt{2}x + 1 = 0}$

$\it{\boxed{5}. \:\:\alpha + \beta = \frac{1}{3}}\\ \\ \it{\alpha * \beta = 1}$

$\it{x^2 - (\frac{1}{3})x + (1) = 0}$

$\it{3x^2 - 1x + 3 = 0}$

$\it{\boxed{6}.\:\: \alpha + \beta = \frac{1}{2}}\\ \\ \it{\alpha * \beta = -4}$

$\it{x^2 - (\frac{1}{2})x + (-4) = 0}$

$\it{2x^2 - 1x - 4 = 0}$

$\it{\boxed{7}.\:\:\alpha + \beta = \frac{1}{3}}\\ \\ \it{\alpha * \beta = \frac{-1}{3}}$

$\it{x^2 - (\frac{1}{3})x + (\frac{-1}{3})}$

$\it{3x^2 - 1x - 1 = 0}$

$\it{\boxed{8}. \:\: \alpha + \beta = \sqrt{3}}\\ \\ \it{\alpha * \beta = 2}$

$\it{x^2 - (\sqrt{3})x + 2 = 0}$

$\it{x^2 - \sqrt{3}x + 2 = 0}$

$\large{\mathbf{\underline{I \: \: HOPE \: \: ITS \: \: HELP \: \: YOU \: \: DEAR, \: \: THANKS}}}$