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Co-ordinate Geometry

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Source: — Problem Solving |

by Rahul@gurome » Fri Jul 09, 2010 8:58 pm
On x-axis, y = 0
If (b^2 - 4ac) > 0 for the equation ax^2 + bx + c = 0, then it gives 2 real roots, and intersects the x-axis in 2 points.

(A) x^2+2x+3 = 0 implies b^2 - 4ac = 4 -12 < 0, implies 2 conjugate imaginary roots, which implies it does not intersect x -axis.
(B) 2x^2-3x+6 = 0 implies b^2 - 4ac = 9 - 48 < 0, implies 2 conjugate imaginary roots, which implies it does not intersect x -axis.
(C) -3x^2-5x+7 = 0 implies b^2 - 4ac = 25 + 84 > 0, which implies equation has 2 real roots, implies it intersects the x -axis in 2 points.

[spoiler]The correct answer is (C).[/spoiler]
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by blaster » Sat Jul 10, 2010 1:26 am
guessing method

as we know that , if an equation contains x^2 ,then the graph must be parabola. and parabola intersects with axis 2 times ,depending on the sign ( it can be also that , parabola will situate above the axis). as all equation are parabola , then we must seek something ,that is differing only one equation from others (especially sign,because mostly it depends on this the draw of parabola).

all this equations are the same ,almost. and you can't simplify them.
(A) y=x^2+2x+3
(B) y=2x^2-3x+6
(D) y=2x^2-4x+2
(E) y=4x^2+2x+1

only C ( y=-3x^2-5x+7) has a different sign of x.

by this method it's easy find an answer, within seconds.
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by GMATGuruNY » Sat Jul 10, 2010 4:57 am
mehulv wrote:33. The graph of which of the following equations have two intersections with axis-x?
(A) y=x^2+2x+3
(B) y=2x^2-3x+6
(C) y=-3x^2-5x+7
(D) y=2x^2-4x+2
(E) y=4x^2+2x+1
The graphing of quadratic equations (parabolas) is not tested on the GMAT, so you shouldn't worry about this type of problem.
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