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If the quadratic equation x2 - (k + 2)x + 4k = 0 has two different integer roots, how many k’s are there?

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If the quadratic equation x2 - (k + 2)x + 4k = 0 has two different integer roots, how many k’s are there?

by Max@Math Revolution » Mon May 11, 2020 6:49 pm

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[GMAT math practice question]

If the quadratic equation x^2 - (k + 2)x + 4k = 0 has two different integer roots, how many k’s are there?

A. 1
B. 2
C. 3
D. 4
E. 5
Last edited by Max@Math Revolution on Fri May 15, 2020 6:22 am, edited 2 times in total.
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Re: In the figure, what is BD^2 - CD^2?

by Max@Math Revolution » Wed May 13, 2020 2:23 am
=>

Assume m and n are integer roots of the equation x^2 - (k + 2)x + 4k = 0.
Then (x - m)(x - n) = x^2 – (m+n)x + mn = x^2 - (k + 2)x + 4k = 0.
Then we have m + n = k + 2 or k = m + n - 2, and mn = 4k.
We have mn = 4k = 4(m + n - 2) or mn = 4m + 4n - 8.
Then, mn – 4m – 4n – 8 = 0 or (m - 4)(n - 4) = 8.
Since m and n are integers, m - 4, and n - 4 are integers and possible solutions of (m-4, n-4) are (1, 8), (2, 4), (4, 2), (8, 1), (-1, -8), (-2, -4), (-4, -2), (-8,- 1).
Then possible solutions of (m, n) are (5, 12), (6, 8), (8, 6), (12, 5), (3, -4), (2, 0), (0, 2) and (-4, 3).

Since the equation has two different roots, we have its discriminant (k + 2)2 -16k = k^2 – 12k + 4 > 0.
Since the roots of k^2 – 12k + 4 = 0 are 6 ± √32, we have k < 6 - √32 or k > 6 + √32.

Since k < 6 - √32 or k > 6 + √32, and k = mn/4, we have possible values of k = mn/4, 15, 12, 12, 15, -3, 0, 0, -3 for (5, 12), (6, 8), (8, 6), (12, 5), (3, -4), (2, 0), (0, 2) and (-4, 3) of (m, n).
Thus, the possible values of k are 15, 12, -3, and 0. We have 4 values.

Therefore, D is the answer.
Answer: D
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