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

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Senior Manager
Joined: 20 Feb 2006
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How many intersects with x-axis does y=x^2+2qx+r have I) [#permalink]  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
VP
Joined: 25 Jun 2006
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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.
Manager
Joined: 12 Jul 2006
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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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Senior Manager
Joined: 20 Feb 2006
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Kudos [?]: 7 [0], given: 0

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?
Senior Manager
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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 â€“ 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?
VP
Joined: 25 Jun 2006
Posts: 1173
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Kudos [?]: 79 [0], given: 0

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 â€“ 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.
Senior Manager
Joined: 20 Feb 2006
Posts: 373
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Kudos [?]: 7 [0], given: 0

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?
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