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Quadratic querie

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by uptowngirl92 » Thu Oct 29, 2009 1:36 am
If f(x)=5x^2 and g(x)=x^2 + 12x + 85, what is the sum of all values for k such that f (k+2)=g(2k) ?

Looks pretty simple..however I got stuck mid-way.
Got till K^2-4K-65=0
Does anyone know how to proceed further??
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by NikolayZ » Thu Oct 29, 2009 1:47 am
Hey again !
You need to use quadratic formula for discriminant, to solve the quadratic equation without factoring.

D=b^2-4ac,
And roots of the equation will be
x12=(-b±sqrt(b^2-4ac))/2a.
D=4+4*65=264
x1=(4+sqrt(264))/2, x2=(4-sqrt(264))/2.
You need to find the sum of the roots, so

x1+x2= (4+sqrt(264)+4-sqrt(264))/2 ==> (4+4)/2=4

Hope this helps.
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Re: Quadratic querie

by Brent@GMATPrepNow » Sun Nov 01, 2009 2:32 pm
uptowngirl92 wrote:If f(x)=5x^2 and g(x)=x^2 + 12x + 85, what is the sum of all values for k such that f (k+2)=g(2k) ?

Looks pretty simple..however I got stuck mid-way.
Got till K^2-4K-65=0
Does anyone know how to proceed further??
You're almost there.
There's a nice rule that says, "If m and n are solutions to the equation x^2 + bx + c = 0, then m+n = -b

In the example k^2-4k-65=0, b=-4
So, m+n = -(-4) = 4
Brent Hanneson - Creator of GMATPrepNow.com
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by NikolayZ » Sun Nov 01, 2009 11:55 pm
Brent ! great one. It is really discriminant terminating thing :)

But we have to be aware of the form of qudratic equation.
Equation must not have an "a" coefficient (a*x^2+bx+c) for this rule to work. I suppose.
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