My if statement is asking that if h is negative, you must do that operation if not the other but do the opposite, do not I understand?
t=b4
e=b3
f=b2
h=e^2-4*t*f
if h==-h
qc=3
p=(h)*(-1)
x5=(sqrt(h))/(2*t);
x6=(-e/(2*t));
x7=(-sqrt(h))/(2*t);
x8=(-h/(2*t));
else
qd=4
x5=(-e+sqrt(h))/(2*t);
x6=(-e-sqrt(h))/(2*t);
end

 Akzeptierte Antwort

David Goodmanson
David Goodmanson am 8 Nov. 2017

0 Stimmen

Hi Erwin
The test h==-h is not a test for whether h is negative. This test is the same as 2*h==0 and always fails unless h = 0. Take a look at h<0, or h<=0 if you intend the condition to work for h=0 as well as negative h.

3 Kommentare

Erwin Avendaño
Erwin Avendaño am 8 Nov. 2017
haha you are right I had forgotten that for positives and negatives it was greater or less than ¡THX!
Erwin Avendaño
Erwin Avendaño am 8 Nov. 2017
Is these two conditions okay? if (qb==2 && qd==4)
David Goodmanson
David Goodmanson am 8 Nov. 2017
They are the right syntax anyway, assuming that qb and qd are scalars. I tend to just use & in every situation because it always works and I don't think it has slowed things down for what I ordinarily do.

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Kategorien

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by