How many intersects with x-axis does y=x^2+2qx+r have I) : Quant Question Archive [LOCKED]
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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]

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08 Sep 2006, 00:48
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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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09 Sep 2006, 13:57
Intersects with X axis -- implies y = 0
We need to find the "real roots" of X.

This becomes a quadratic eqn with solutions given by:

X = -2q + sqrt(q^2 - r) and -2q - sqrt(q^2 -r)

Real roots are only possible if the radical is positive, i.e: q^2 -r > 0
(Note: complex numbers are out of scope in GMAT)

From 1:
q^2 > r, which meets are condition, Hence SUFF

From 2:
Not possible to determine relationship between q^2 and r, hence INSUFF

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09 Sep 2006, 16:55
getting A too, same method as above.
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10 Sep 2006, 16:42
Can someone explain how you arrived arrived at the solution for the quadratic? Keep it simple please brainiacs
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10 Sep 2006, 17:43
Solution of a quadratic equation ax^2 + bx + c is

X = (-b +/- sqrt(b^2 â€“ 4ac))/2a
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10 Sep 2006, 18:48
Can someone explain how you arrived arrived at the solution for the quadratic? Keep it simple please brainiacs

Let me try...

The question can be rephrased as:

For how many real values of x can y be 0. In other words, we have to find out if the equation x^2+2qx+r = 0 has real roots, and if there are real roots, then how many are there?

A quadratic equation can have 0, 1 or 2 real roots....

Lets see how this is possible:

The formula for roots of a quadratic is given by:

x1 = [-b + sqrt(b^2-4ac)]/2a
x2 = [-b - sqrt(b^2-4ac)]/2a

As can be seen there are 0 real roots if b^2-4ac is -ve, because then each root includes an imaginary number.

There are 2 real roots if b^2-4ac is +ve

There is 1 real root if b^2-4ac = 0

With this knowledge, one should be able to solve the problem.
10 Sep 2006, 18:48
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