Question

The sum and product of the zeroes of a quadratic polynomial are − 19 /20 and − 3 /10 , respectively. If the coefficient of x^ 2 is 20 , find the quadratic polynomial.

Answer 1

Given :

• Sum of zeroes = -19/20
• Product of zeroes = -3/10

To find :

The quadratic polynomial.

Theory :

if $\sf\alpha \: and \beta$ are the zeros of a quadratic polynomial

f(x) . Then the polynomial f x is given by

$\sf \:f(x) = k(x {}^{2} - ( \alpha + \beta )x + \alpha \beta )$

or

$\sf \: f(x) = k(x {}^{2} - (sum \: of \: the \: zeroes)x + product \: of \: the \: zeroes)$

Solution :

Let $\sf \alpha \: and \: \beta$ are the zeros of the required polynomial.It is given that

$\sf \alpha + \beta = \frac{-19}{20}$

$\sf \: and \: \alpha \beta = \frac{-3}{10}$

The quadratic polynomial is

$\sf \: f(x) =k( x {}^{2} - ( \alpha + \beta )x + \alpha \beta )$

$\sf \implies \: f(x) =k( x {}^{2} - (\frac{-19}{20}) + \frac{-3}{10})$

Taking the LCM:-

$\sf\implies\:f(x)=k(\dfrac{20x^{2}+19x-6}{20})$

Answer 2

Answer:

Given :

Sum of zeroes = -19/20

Product of zeroes = -3/10

To find :

The quadratic polynomial.

Theory :

if \sf\alpha \: and \betaαandβ are the zeros of a quadratic polynomial

f(x) . Then the polynomial f x is given by

\sf \:f(x) = k(x {}^{2} - ( \alpha + \beta )x + \alpha \beta )f(x)=k(x

2

−(α+β)x+αβ)

or

\sf \: f(x) = k(x {}^{2} - (sum \: of \: the \: zeroes)x + product \: of \: the \: zeroes)f(x)=k(x

2

−(sumofthezeroes)x+productofthezeroes)

Solution :

Let \sf \alpha \: and \: \betaαandβ are the zeros of the required polynomial.It is given that

\sf \alpha + \beta = \frac{-19}{20}α+β=

20

−19

\sf \: and \: \alpha \beta = \frac{-3}{10}andαβ=

10

−3

The quadratic polynomial is

\sf \: f(x) =k( x {}^{2} - ( \alpha + \beta )x + \alpha \beta )f(x)=k(x

2

−(α+β)x+αβ)

\sf \implies \: f(x) =k( x {}^{2} - (\frac{-19}{20}) + \frac{-3}{10})⟹f(x)=k(x

2

−(

20

−19

)+

10

−3

)

Taking the LCM:-

\sf\implies\:f(x)=k(\dfrac{20x^{2}+19x-6}{20})⟹f(x)=k(

20

20x

2

+19x−6

)