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(x-6)(x+k)=x^2-4kx+m/5

In the equation above, k and m are constants. If the equation is true for all values of x, what is the value of m?

A)0

B)4

C)--16

D)-18/5

E)--36

Nina,

First off, notice that 6 and -k are roots of the eqn, i.e., x^2-4kx+m/5 = 0 for both 6 and -k. Use this result to solve for k in terms of m by plugging k into the quadratic, which gives

k^2 + 4k^2 + m/5 = 0

so that k = sqrt(-m)/5. For convenience, let z = sqrt(-m) (note that this means m < 0 for z real), then the eqn becomes

z^2 +24z -180 = 0.

Using the familiar quadratic eqn, i.e., for Az^+Bz+C = 0, z = (-B+/-sqrt(B^2-4AC))/2A, we have

z = -24/2+/- sqrt(24^2+4x180)/2 = -12 +/- 18.

Using -12+18 = 6, we have m = -z^2 = -36, or choice E.

BTW, I assume that the "--" symbol really means "-".

Hope this makes sense.

Randy

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