Question

Find the quadroht polynomial whose zeros are 3+√5/3 and 3-√5/3​

Answer 1

Answer :

The required quadratic polynomial is

 $\bf x^2-6x+\frac{76}{9}$

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 zeroes of the quadratic polynomial,

    $\bf 3+\frac{\sqrt{5}}{3} \ \ and \ \ 3-\frac{\sqrt{5}}{3}$

⇒ Sum of zeroes

              $= 3+\frac{\sqrt{5}}{3} + 3-\frac{\sqrt{5}}{3} \\\\ =3+3 \\\\ =6$

⇒ Product of zeroes

              $=( 3+\frac{\sqrt{5}}{3} )(3-\frac{\sqrt{5}}{3}) \\\\ =3(3-\frac{\sqrt{5}}{3}) +\frac{\sqrt{5}}{3}(3-\frac{\sqrt{5}}{3}) \\\\ =9-\sqrt{5}+\sqrt{5}-\frac{5}{9} \\\\ =9-\frac{5}{9} \\\\ =\frac{81-5}{9} \\\\ =\frac{76}{9}$

The quadratic polynomial is of the form

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

$=> x^2-(6)x +\frac{76}{9} \\\\ \bf => x^2-6x+\frac{76}{9}$

∴ The required quadratic polynomial is

        $\bf x^2-6x+\frac{76}{9}$