Question

establish the relation between roots and different coefficient in a quadratic equation​

Answer 1

Quadratic equation :

• Quadratic equation is the equation of a polynomial having highest degree of the variable as 2.

• General form : ax² + bx + c = 0

  where

a is the coefficient of x²

b is the coefficient of x

c is the constant term

• The roots/zeroes of the quadratic equation are the solutions of the equation which are the values of the variable x satisfying the equation.

 we know,

quadratic formula :

   $\sf x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$

Let α and β are the roots of the quadratic equation ax² + bx + c = 0

 $\sf \alpha =\dfrac{-b + \sqrt{b^2-4ac}}{2a} \\\\ \sf \beta=\dfrac{-b - \sqrt{b^2-4ac}}{2a}$

RELATION BETWEEN ROOTS AND COEFFICIENTS :

• Sum of roots = α + β

$\sf \alpha + \beta =\dfrac{-b + \sqrt{b^2-4ac}}{2a}+\dfrac{-b - \sqrt{b^2-4ac}}{2a} \\\\ \sf \alpha + \beta =\dfrac{-b + \sqrt{b^2-4ac}-b- \sqrt{b^2-4ac}}{2a} \\\\ \sf \alpha + \beta =\dfrac{-b-b}{2a} \\\\ \sf \alpha + \beta =\dfrac{-2b}{2a}\\\\ \boxed{\tt \alpha + \beta =\dfrac{-b}{a}=\dfrac{-(x \ coefficient)}{x^2 \ coefficient}}$

• Product of roots = αβ

$\sf \alpha \beta =\bigg(\dfrac{-b + \sqrt{b^2-4ac}}{2a}\bigg) \times \bigg(\dfrac{-b - \sqrt{b^2-4ac}}{2a} \bigg)\\\\ \sf \alpha \beta =\dfrac{(-b)^2 - (\sqrt{b^2-4ac})^2}{(2a)^2} \\\\ \sf \alpha \beta =\dfrac{b^2-(b^2-4ac)}{4a^2} \\\\ \sf \alpha \beta =\dfrac{b^2-b^2+4ac}{4a^2} \\\\ \sf \alpha \beta =\dfrac{4ac}{4a^2} \\\\ \boxed{\tt \alpha \beta =\dfrac{c}{a}=\dfrac{constant \ term}{x^2 \ coefficient}}$

Therefore,

Sum of roots = -(x coefficient)/x² coefficient

Product of roots = constant term/x² coefficient