Question

If two roots of quadratic equation are, -2/3 and -3/5 , then quadratic equation will be :

Answer 1

Answer :

$x^{2} +\frac{19}{15} x+\frac{2}{5}$

Step-by-step explanation :

Quadratic Polynomials :

        ✯ It is a polynomial of degree 2

        ✯ General form :

                  ax² + bx + c  = 0

                $\boxed{\bf x=\frac{-b\pm\sqrt{b^2-4ac} }{2a} }$            

        ✯ Determinant, D = b² - 4ac

        ✯ Based on the value of Determinant, we can define the nature of roots.

                D > 0 ; real and unequal roots

                D = 0 ; real and equal roots

                D < 0 ; no real roots i.e., imaginary

        ✯ Relationship between zeroes and coefficients :

                  ✩ Sum of zeroes = -b/a

                  ✩ Product of zeroes = c/a

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       Given,

two roots of quadratic equation are, -2/3 and -3/5

$\bigstar$ Sum of zeroes

             $=\frac{-2}{3} +\frac{-3}{5} \\\\ =\frac{-2}{3} -\frac{3}{5} \\\\ =\frac{-2(5)-3(3)}{3\times5} \\\\ =\frac{-10-9}{15} \\\\ =\frac{-19}{15}$

$\bigstar$ Product of zeroes

              $=(\frac{-2}{3})(\frac{-3}{5}) \\\\ =\frac{6}{15} \\\\ =\frac{2}{5}$

The quadratic polynomial will be in the form of

=> x² - (sum of zeroes)x + (product of zeroes)

=> x² - (-19/15)x + (2/5)

=> x² + (19/15)x + 2/5

Answer 2

Step-by-step explanation:

refer the attachment..........