UsukiDoll
  • UsukiDoll
Consider the quadratic equation ax^2+bx+c. Suppose that b and c can only take the values [1,2,3,4,5,6]. For which values of b and c does the equation have real roots?
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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UsukiDoll
  • UsukiDoll
do I have to use the quadratic formula |dw:1377668124007:dw|on each combination ?
UsukiDoll
  • UsukiDoll
alright..but do I have to plug in the values to the discriminant?
terenzreignz
  • terenzreignz
Whether or not it has real roots also depends on the value of a, right? I mean, if a is negative, then it doesn't really matter what b and c are, the quadratic expression WILL have real roots...

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More answers

UsukiDoll
  • UsukiDoll
ummm how did you get b= c/2?
anonymous
  • anonymous
look at the discriminant b^2-4ac putting the values of abc u can take ur answer
UsukiDoll
  • UsukiDoll
ok if we let b=2 and c = 1 b^2-4ac 2^2-4(1)(1) = 4-4 =0
anonymous
  • anonymous
here we have to consider a too
UsukiDoll
  • UsukiDoll
but a is like one on default since b and c is 1-6
terenzreignz
  • terenzreignz
Are you absolutely sure that a = 1? Because it doesn't say so... I'm not really a fan of assumptions ^_^
UsukiDoll
  • UsukiDoll
so it has to be something...like a b c values that can produce something greater than 0?
anonymous
  • anonymous
hey poeple a= not equal to one ... for real roots b2-4ac=0 frrom here putting b c we can find a.
UsukiDoll
  • UsukiDoll
ok so a is unknown then.
terenzreignz
  • terenzreignz
Totally unknown (it would seem) but if it's negative, then it doesn't matter which value (from {1,2,3,4,5,6}) b or c takes, it will have real roots... So I guess the interesting question is what if a > 0 ?
anonymous
  • anonymous
ryt Mr.@terenzreignz
UsukiDoll
  • UsukiDoll
meaning what happens if a is greater than zero a.k.a a positive value
UsukiDoll
  • UsukiDoll
it won't have real roots?
terenzreignz
  • terenzreignz
Not really, but it will only have real roots for certain values of b and c, which... we are about to find out...
UsukiDoll
  • UsukiDoll
so if a is positive, not all of the values of b and c will produce a real root. Only some of them.
anonymous
  • anonymous
if a 4 times the product of a c is bigger than b^2 then ur root will the non real :)
anonymous
  • anonymous
oh ok @terenzreignz never mind .. m not gonna call u anyway :P
terenzreignz
  • terenzreignz
Understood. I'm still undecided about a... it has quite a few possibilties.
UsukiDoll
  • UsukiDoll
but how do I show that on b^2-4ac > 0
terenzreignz
  • terenzreignz
Let's explore... for one thing, it means \[\Large b^2 > 4ac\]
terenzreignz
  • terenzreignz
We can take the square root of both sides... \[\Large b > 2\sqrt{ac}\]
terenzreignz
  • terenzreignz
Which is possible, mind you, assuming a > 0.
UsukiDoll
  • UsukiDoll
and if a < 0?
terenzreignz
  • terenzreignz
I told you, if a < 0, then it doesn't matter what values b and c take from {1,2,3,4,5,6} it will *always* have real roots.
UsukiDoll
  • UsukiDoll
andif a > 0 then only certain values of b and c will produce a real root.
terenzreignz
  • terenzreignz
Because if a < 0 and b and c are from {1,2,3,4,5,6} then \[\Large -4a > 0\]\[\Large -4ac>0\]\[\Large b^2 -4ac > 0\]\[\Large \sqrt{b^2-4ac}\in \mathbb{R} \]
terenzreignz
  • terenzreignz
Yes, we need to find them, it would seem.
UsukiDoll
  • UsukiDoll
so how do you find the roots?
terenzreignz
  • terenzreignz
The roots? Or the values of b and c? And I'm working on it...
terenzreignz
  • terenzreignz
All the same, I wouldn't mind some more expert opinion XD
UsukiDoll
  • UsukiDoll
well this is the first time that I've ever dealt with a problem like this x.x.x.x.x
UsukiDoll
  • UsukiDoll
@dumbcow

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