Question

Find the quadratic polynomial whose roots are 2 and -1/3.​

Answer 1

Given

$\tt \to \: Roots \: \: are \: \: 2 \: and \: \dfrac{ - 1}{3}$

To Find the quadratic equation

Now , we know that

$\tt \to \: ( \alpha + \beta ) = 2$

$\tt \to \alpha \beta = \dfrac{ - 1}{3}$

General Equation of Quadratic equation

$\tt \to \: {x}^{2} - ( \alpha + \beta )x + \alpha \beta = 0$

Now Put the Value on formula

$\tt \to \: {x}^{2} - (2)x + \bigg( \dfrac{ - 1}{3} \bigg) = 0$

Now Simplify the equation

$\tt \to \: {x}^{2} - 2x - \dfrac{1}{3} = 0$

$\tt \to \dfrac{3 {x}^{2} - 6x - 1}{3} = 0$

$\tt \to \: 3 {x}^{2} - 6x - 1 = 0$

Answer

$\tt \to \: 3 {x}^{2} - 6x - 1 = 0$

More Information

$\tt \to \: ( \alpha + \beta ) = \dfrac{ - b}{a}$

where b is coefficient of x and a is coefficient of x²

$\tt \to \: ( \alpha \beta ) = \dfrac{c}{a}$

where c is constant and a is coefficient of x²

Quadratic Formula

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