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How many intersects with x-axis does y=x^2+2qx+r have I)

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How many intersects with x-axis does y=x^2+2qx+r have I) [#permalink] New post 30 Nov 2006, 16:09
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How many intersects with x-axis does y=x^2+2qx+r have

I) q^2>r

II) r^2>q
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 [#permalink] New post 30 Nov 2006, 17:34
A.


(2q)^2 - 4r = 4 (q^2 - r)

all u have to do is to determine q^2 - r is positive or negative or 0.
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 [#permalink] New post 30 Nov 2006, 18:29
Sorry - I don't see how you got this equation: (2q)^2-4r, which you simplified to 4(q^2-r). Could you explain further?
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 [#permalink] New post 30 Nov 2006, 18:32
I managed to get the correct answer but if this were a PS question - how would you know if there were 2,1 or 0 x intercepts?
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 [#permalink] New post 30 Nov 2006, 18:35
wshaffer wrote:
Sorry - I don't see how you got this equation: (2q)^2-4r, which you simplified to 4(q^2-r). Could you explain further?


Solution of a quadratic equation ax^2 + bx + c is

X = (-b +/- sqrt(b^2 – 4ac))/2a

In this case, a=1 b=2q and c=r

Substituting into the bold (2q)^2 - 4(1*r) = 4q^2 - 4r = 4(q^2 - r)

If you can determine this value of (2q)^2 - 4(1*r) then you know how many intercepts there are - I think.

Tennisball - care to elaborate?
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 [#permalink] New post 30 Nov 2006, 19:54
MBAlad wrote:
wshaffer wrote:
Sorry - I don't see how you got this equation: (2q)^2-4r, which you simplified to 4(q^2-r). Could you explain further?


Solution of a quadratic equation ax^2 + bx + c is

X = (-b +/- sqrt(b^2 – 4ac))/2a

In this case, a=1 b=2q and c=r

Substituting into the bold (2q)^2 - 4(1*r) = 4q^2 - 4r = 4(q^2 - r)

If you can determine this value of (2q)^2 - 4(1*r) then you know how many intercepts there are - I think.

Tennisball - care to elaborate?


Yes. all you need to know is b^2 - 4ac.
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 [#permalink] New post 01 Dec 2006, 08:45
But if that were a PS equation what would the resulting value tell us? ie how do we know if there are 0,1 or 2 intercepts?
  [#permalink] 01 Dec 2006, 08:45
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