Question

A quadratic polynomial whose zeroes are 3/5 and -1/2 is

Answer 1

Step-by-step explanation:

$let \: \alpha \: and \: \beta \: be \: the \: zeros \: of \: required \: \\ quadratic \: polynomial. \: therefore \\ sum \: of \: zeros \\ \alpha + \beta = \frac{3}{5} + ( - \frac{1}{2} ) \\ = \frac{6 - 5}{10} \\ = \frac{1}{10} \\ product \: of \: zeros \\ \alpha \beta = \frac{3}{5} ( - \frac{1}{2} ) = - \frac{3}{10} \\ let \: the \: required \: polynomial \: be \: \\ p(x) = {x}^{2} - ( \alpha + \beta ) + \alpha \beta \\ = {x}^{2} + \frac{1}{10} x + ( - \frac{3}{10} ) \\ = 10 {x}^{2} + x - 3$

Answer 2

Answer:

letαandβbethezerosofrequired

quadraticpolynomial.therefore

sumofzeros

α+β=

5

3

+(−

2

1

)

=

10

6−5

=

10

1

productofzeros

αβ=

5

3

(−

2

1

)=−

10

3

lettherequiredpolynomialbe

p(x)=x

2

−(α+β)+αβ

=x

2

+

10

1

x+(−

10

3

)

=10x

2

+x−3