Question

If a quadratic polynomial y=ax2+bx+c intersects x axis at alpha and bita, then?

Answer 1

Answer:

alpha = [-b + sqrt(b^2 - 4ac)]/2a

beta  = [-b - sqrt(b^2 - 4ac)]/2a

Step-by-step explanation:

A quadratic equation has a graph of a parabola. If the quadratic polynomial y=ax2+bx+c intersect x axis at two points than , it has two roots .

This is because number of roots of a quadratic polynomial is equal number of points it cuts x axis .

So, here y=ax2+bx+c intersect x axis at two points so, it has two roots alpha and beta .

Value of alpha  =  [-b + sqrt(b^2 - 4ac)]/2a

Value of beta  =  [-b - sqrt(b^2 - 4ac)]/2a

Answer 2

Answer:

$\alpha(\frac{-b+\sqrt{b^2-4ac}}{2a},0)\:and\:\beta(\frac{-b-\sqrt{b^2-4ac}}{2a},0)$

Step-by-step explanation:

Concept:

The roots of the quadratic eqquation ax²+bx+c=0 are given by

$x=\frac{-b+\sqrt{b^2-4ac}}{2a}\:and\:x=\frac{-b-\sqrt{b^2-4ac}}{2a}$

since the polyniomial y=ax²+bx+c intersects x axis, y=0

This implies, ax²+bx+c=0

Then,

$x=\frac{-b+\sqrt{b^2-4ac}}{2a}\:and\:x=\frac{-b-\sqrt{b^2-4ac}}{2a}$

Therefore, the coordinates of $\alpha\:and\:\beta$ are

$(\frac{-b+\sqrt{b^2-4ac}}{2a},0)\:and\:(\frac{-b-\sqrt{b^2-4ac}}{2a},0)$