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# What is the highest integral value of 'k' for which the quad

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Joined: 10 Jul 2013
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What is the highest integral value of 'k' for which the quad  [#permalink]

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22 Aug 2013, 11:43
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Difficulty:

45% (medium)

Question Stats:

57% (00:48) correct 43% (01:00) wrong based on 167 sessions

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What is the highest integral value of 'k' for which the quadratic equation x^2 - 6x + k = 0 have two real and distinct roots?

A. 9
B. 7
C. 3
D. 8
E. 12

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Re: What is the highest integral value of 'k' for which the quad  [#permalink]

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22 Aug 2013, 11:53
Asifpirlo wrote:
What is the highest integral value of 'k' for which the quadratic equation x^2 - 6x + k = 0 have two real and distinct roots?

A. 9
B. 7
C. 3
D. 8
E. 12

for a quadratic equation $$ax^2+bx+c=0$$
to have 2 different roots discriminant D must be greater than 0.
$$D =b^2-4ac$$
hence $$36-4k>0$$
$$k<9$$
so highest integral value = $$8$$
hence D
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Re: What is the highest integral value of 'k' for which the quad  [#permalink]

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22 Aug 2013, 23:02
for any quadratic equation $$ax^2 + bx + c = 0$$, the nature of its roots can be determined by knowing its discriminant(D). $$Discriminant = b^2 - 4ac$$

if Discriminant = 0, then Quadratic equation will have real and equal roots.
if Discriminant > 0, then Quadratic equation will have real and unequal roots.
if Discriminant < 0, then Quadratic equation will have imaginary roots.

in our case for $$x^2 - 6x + k = 0$$ to have real and distinct roots, $$(-6)^2 - 4k$$ must be greater than zero. So we have $$(-6)^2 - 4k > 0$$ ----------> $$36 - 4k > 0$$ --------------> $$4k < 36$$ ---------> $$k < 9$$ ------------> The greatest value of K will be 8.
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Re: What is the highest integral value of 'k' for which the quad  [#permalink]

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18 Feb 2017, 11:52
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