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Intern
Joined: 12 Mar 2018
Posts: 4

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12 Mar 2018, 05:51
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What is the largest integral value of 'k' for which the quadratic equation x2 - 6x + k = 0 will have two real and distinct roots?

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VP
Joined: 30 Jan 2016
Posts: 1158

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Updated on: 13 Mar 2018, 16:32
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Using discriminant:

$$x1,2=\frac{6+\sqrt{36-4k}}{2}$$=$$\frac{6+\sqrt{4*(9-k)}}{2}$$=$$\frac{6+2\sqrt{(9-k)}}{2}$$=$${3+\sqrt{(9-k)}}$$. So, for the expression to have distinct real roots, $$\sqrt{9-k}$$ has to be an integer, therefore k can:

k=5, x1=$${3+\sqrt{(9-5)}}=3+2=5$$, or x2=$${3+\sqrt{(9-5)}}=3-2=1$$
k=8, x1=$${3+\sqrt{(9-8)}}=3+1=4$$, or x2=$${3+\sqrt{(9-5)}}=3-1=2$$
k=9, x=$${3+\sqrt{(9-9)}}=3$$

So, the expression becomes $$x^2-6x+9=(x-3)^2=0$$
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Non progredi est regredi

Originally posted by Akela on 12 Mar 2018, 11:48.
Last edited by Akela on 13 Mar 2018, 16:32, edited 7 times in total.
VP
Joined: 30 Jan 2016
Posts: 1158

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13 Mar 2018, 14:51
I've made a mistake in my previous post. The answer is 9 using Vieta's formula:

$$x^2-6x+k=0$$
x1+x2=6
x1*x2=k

x1 and x2 must be > 0 to maximize k and x1 and x2 cannot have different signs:

x1=1, x2=5 => x1*x2=k=5
x1=2, x2=4 => x1*x2=k=8
x1=3, x2=3 => x1*x2=k=9
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Non progredi est regredi
Math Expert
Joined: 02 Sep 2009
Posts: 58340

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17 Mar 2018, 08:20
nshtkum wrote:
What is the largest integral value of 'k' for which the quadratic equation x2 - 6x + k = 0 will have two real and distinct roots?

--== Message from the GMAT Club Team ==--

THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION.
This discussion does not meet community quality standards. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.

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