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# If (x +9) is one of the factors of x2 + mx 4c = 0 and m and c are

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Re: If (x +9) is one of the factors of x2 + mx 4c = 0 and m and c are [#permalink]
$$x^2+mx-4c=0$$
we are told that (x+9) is a root, implying that x=-9 gives us 0 in the above equation.
$$(-9)^2-9m-4c=0$$
$$81-9m-4c=0$$
$$c=\frac{9(-9+m)}{-4}$$
Notice that we are told that c and m are both positive integers.
As the fraction is divided by a negative number, this places the restriction on the value of m:
(-9+m)<0
0<m<9.
M must be less than 9 and greater than 0.
As 4 does not divide 9, (-9+m) must be some multiple of 4.
The first case is m=1, which gives us c=9*8/4=9*2=18.
mc in this case would be 18. This is not an option, so there must another case
The second case is m=5. This gives us c=9*4/4=9
mc=45, which is the option E and is therefore the correct answer.
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Re: If (x +9) is one of the factors of x2 + mx 4c = 0 and m and c are [#permalink]
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Re: If (x +9) is one of the factors of x2 + mx 4c = 0 and m and c are [#permalink]
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