When does a quadratic equation in the form ax2+b=c have two real rational solutions?

Hello from MrBillDoesMath!

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

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

step-by-step explanation:

abc