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#1

$Old$ July 4th, 2006, 11:44 AM

bobby77 $bobby77 is offline$

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$Default$ Intersection of curves

The line y=2x- 3 intersects the curve y=x^2+x-15 at the points

(a)(3,3)and(4,5)
(b)(-3,-9)and(4,5)
(c)(3,3)and(-4,-11)
(d)(-3,-9)and(-4,-11).

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#2

$Old$ July 4th, 2006, 12:21 PM

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Quote:

Originally Posted by bobby77

The line y=2x- 3 intersects the curve y=x^2+x-15 at the points

(a)(3,3)and(4,5)
(b)(-3,-9)and(4,5)
(c)(3,3)and(-4,-11)
(d)(-3,-9)and(-4,-11).

At a point of intersection:

$2x-3=x^2+x-15$ ,

so:

$x^2-x-12=0$

Now trying the candidates for $x$ , we find that , $x=-3$ satisfies
this equation as does $x=4$ , so (b) must be the answer (assuming the
$y$ 's are OK which I suppose we can).

RonL

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