how do you explain how two different equations have real rational and real irrational solutions? the equations are f(x) = x^2+6x-16 and
g(x) = x^2 + 6x + 1

Question

how do you explain how two different equations have real rational and real irrational solutions? the equations are f(x) = x^2+6x-16 and 
g(x) = x^2 + 6x + 1

jhannybean · Answer

discriminant: $d= b^2-4ac $

$d > 0$ : 2 real and unreal roots

$d =0$ : only 1 real root.

$d <0$ : pair of complex roots in the form $a\pm bi$ where $i^2=-1$

jhannybean · Answer

oh and $d >0$ for perfect squares means you will have 2 rational roots.

jhannybean · Answer

Both of your equations are in quadratic form: $ax^2+bx+c=0$, thus you have a discriminant: $b^2-4ac$. You can use your discriminant to find your roots, to see whether they are complex or rational.

anonymous · Answer

thank you so much !