Question

Form the Quadratic polynomials whose zeros are:3 and 1/5

Answer 1

Answer:

15x^2-x+5

Step-by-step explanation:

x^2(a+b)-(ab)+1=0

3x^2-1/5x+1=0

(15x^2-x+5)/5=0

15x^2-x+5=0

Answer 2

Question:

Form the Quadratic polynomials whose zeros are:

$3 \: and \: \frac{1}{5}$

Answer:

$The \: equation \: is,$

$\underline{ \: \sf \red{ \bold{5 {x}^{2} - 16x + 3 = 0}} \: }$

Given:

$\bigstar Zeroes \: of \: polynomial \: are: \: 3 \: and \: \frac{1}{5}$

Step-by-step explanation:

The equation is,

$\boxed{ {x}^{2} - ( \alpha + \beta )x + \alpha \beta = 0}$

$Here, \: \alpha = 3, \: \beta = \frac{1}{5}$

$\implies {x}^{2} - ( 3+ \frac{1}{5} )x + (3 \times \frac{1}{5} ) = 0$

$\implies {x}^{2} - ( \frac{16}{5} )x + (\frac{3}{5}) = 0$

Now multiplying by 5 on both sides,

$\implies \underline{ \: \boxed{5 {x}^{2} - 16x + 3 = 0} \: }$

$\therefore The \: required \: equation \: is,$

$\underline{ \: \sf \pink{5 {x}^{2} - 16x + 3 = 0} \: }$

More information:

$\bigstar Sum\: of\:roots\:(\alpha+\beta)=\frac{-b}{a}$

$\bigstar Product\: of\: roots\: (\alpha\beta)=\frac{c}{a}$