Question

form quadratic polynomial whose zeroes are 3 and 1/5​

Answer 1

Answer:

1/5 (5x^2-16x+3)

Step-by-step explanation:

As zeroes are 3,1/5

(x-3) & (x-1/5) are factors of the polynomial

So the polynomial is (x-3)*(x-1/5)

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

=x^2-16/5x+3/5

or

1/5 (5x^2-16x+3)

Answer 2

Answer:

$5x^{2} -16x+3=0$  is the required quadratic polynomial.

Step-by-step explanation:

If $\alpha$ and $\beta$ are the roots of any quadratic polynomial than quadratic polynomial can be expressed as

$Quadratic-Polynomial =$ $(x-\alpha ) (x-\beta )$

$(Quadratic-Polynomial) = x^{2} -(\alpha +\beta )x+\alpha \beta$

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

Required Polynomial $= (x-3)(x-\frac{1}{5} )$

                                  $=x^{2} -\frac{1}{5} x-3x+\frac{3}{5}$

                                  $=x^{2} -\frac{16x}{5}+\frac{3}{5}$

                                $=\frac{1}{5} (5x^{2} -16x+3)$

      $\frac{1}{5} (5x^{2} -16x+3)$  which is the required quadratic polynomial.  

   

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