Question

If a, b, c ∈ R, a > 0,c< 0, then prove that the roots of ax²+bx+c=0 are real and distinct.

Answer 1

Dear Student ,

Solution:

As a > 0 , c < 0

So we can write the equation
$a {x}^{2} + bx + c = 0$
as

$a {x}^{2} + bx - c = 0$
Now, roots of equation can be calculated by Quadratic formula

$x = \frac{ - b + - \sqrt{ {b}^{2} - 4ac } }{2a}$
from the above equation put c = -c

So,we get
$x \: = \frac{ - b + - \sqrt{ {b}^{2} + 4ac } }{2a}$
So,
$\sqrt{ {b}^{2} + 4ac}$
gives positive value always

that is both the roots of equation are real and distinct ,and these are
$x1 = \frac{ - b + \sqrt{ {b}^{2} + 4ac } }{2a} \\ \\ x1 = \frac{ - b - \sqrt{ {b}^{2} + 4ac } }{2a}$
Hence prove.

Hope it helps you.

Answer 2

Hi ,

It is given that ,

a , b , c € R , a > 0 , c > 0 ,


Quadratic equation ax² + bx + c = 0

discriminant = ∆

∆ = b² - 4ac

= b² - 4 × a × ( - c ) [ since c < 0 ]

= b² + 4ac

> 0

∆ > 0

Therefore ,

roots are real and distinct.

I hope this helps you.

: )