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by gmatmachoman » Fri May 28, 2010 3:20 am
What is the highest integral value of 'k' for which the quadratic equation x2 - 6x + k = 0 have two real and
distinct roots?

A. 9
B. 7
C. 3
D. 8
E. 12
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by gmatjedi » Fri May 28, 2010 3:26 am
equation needs to be in form of (x-a)(x-b)

d: (x-4)(x-2)
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by selango » Fri May 28, 2010 3:29 am
Substitute options in the equation,

x2-6x+k=0

Only 9 and 8 satisfies the condition.

From this only 8 has 2 distinct roots

So 8 is answer
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by indiantiger » Fri May 28, 2010 3:33 am
Here is my brute force method, replace the value of k by answer choices:
A. 9 => x^2 - 6x + 9=> (x-3)^2 NO
B. 7 => x^2 - 6x + 7 => NO
C. 3 => x^2-6x +3 => NO
D. 8 => x^2 -6x + 8 => x^2 -4x-2x+8 => (x-4)(x-2) {ANSWER}
E. 12 => x^2 -6x +12 =>NO
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by liferocks » Fri May 28, 2010 4:20 am
for x2 - 6x + k = 0 have real and distinct root
36-4k>0
or 9>k
so highest integral value of k=8

Ans option D
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by indiantiger » Fri May 28, 2010 2:44 pm
@LifeRocks

awesome solution :), never thought of using b^2 - 4ac
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