What is true about the solutions of a quadratic equation when the radicand of the quadratic formula is a perfect square?
Answer
No real solutions
Two identical rational solutions
Two different rational solutions
Two irrational solutions
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OpenStudy (anonymous):
Two different rational solutions
OpenStudy (anonymous):
Two identical rational solutions.
OpenStudy (cwrw238):
isnt 2 identical myko?
OpenStudy (anonymous):
If 'identical' means +/-
OpenStudy (tennistar):
um I have two answers now
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OpenStudy (anonymous):
\[x=-b \pm \sqrt{b ^{2}-4ac}/2a\]
you will get -b +- n/2a, so two different rational solutions
OpenStudy (cwrw238):
like for example (x - 2)^2 = 0 ?
OpenStudy (anonymous):
@cwrw238 yeah that's where I was going too :)
OpenStudy (tennistar):
So c is the answer
OpenStudy (anonymous):
but here says radicand in the cuadratic formula, not that the equation is perfect square
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OpenStudy (anonymous):
Two different rational solutions
i insist on this
OpenStudy (cwrw238):
oh radicand??
OpenStudy (anonymous):
radicand of the quadratic formula is a rather strange way to say "discriminant" aka \(b^2-4ac\)
OpenStudy (cwrw238):
- not sure npw lol!!
OpenStudy (anonymous):
if it is a perfect square then there are two rational zeros
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OpenStudy (anonymous):
@myko is right
OpenStudy (anonymous):
unless of course \(b^2-4ac=0\) which is also a perfect square. in that case there is one rational zero
OpenStudy (tennistar):
ok thanks everyone
OpenStudy (cwrw238):
yes - i misread the question - but the radicand thing confused me