Hello from MrBillDoesMath!
Answer:
The roots are real and rational when b^2+4ac > 0
Discussion:
I suspect the quadratic equation you meant was
ax^2 + bx = c (*)
Note: the equation in the Question contains only "b", not "bx.
Rewriting (*) as
ax^2 + bx + (-c) = 0
and using the quadratic equation gives
x = ( -b +\- sqrt(b^2-4a(-c) ) /(2a)
= ( -b +\- sqrt(b^2 +4ac) )/(2a)
So the roots (values of x) are real and rational when b^2+4ac > 0 (in this case sqrt(b^2 +4ac) is a real number)
Thank you,
MrB
The equation given in the question is [tex]ax^2+bx=c[/tex]. Thus, if the value of [tex]b^2+4ac[/tex] is greater than zero then the equation always gives real roots.
If the discriminant of any quadratic equation is greater than zero, then the roots of the equation are always real.
The formula for finding the roots of a quadratic equation that has the equation [tex]ax^2+bx+c=0[/tex] is given as:
[tex]D=b^2-4ac[/tex]
Thus apply the formula for the given equation and solve it further.
For real roots, discriminant must be greater than zero.
[tex]D>0\\b^2-4a(-c)>0\\b^2+4ac>0[/tex]
Thus, if the value of [tex]b^2+4ac[/tex] is greater than zero then the equation always gives real roots.
To know more about quadratic equations, please refer to the link.
https://brainly.com/question/12186700
