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In the equation x^2+kx+1=0, x is a variable and k is a constant

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In the equation x^2+kx+1=0, x is a variable and k is a constant [#permalink]

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New post 10 Aug 2015, 16:53
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In the equation \(x^2+kx+1=0\), x is a variable and k is a constant. If the quadratic equation has two distinct real roots, which of the following could be true?

I) k=0

II) k=−1

III) k=−3

A) I only
B) III only
C) I and II only
D) II and III only
E) I and III only
[Reveal] Spoiler: OA

Last edited by ENGRTOMBA2018 on 10 Aug 2015, 17:03, edited 1 time in total.
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Re: In the equation x^2+kx+1=0, x is a variable and k is a constant [#permalink]

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New post 10 Aug 2015, 17:05
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Mascarfi wrote:
In the equation \(x^2+kx+1=0\), x is a variable and k is a constant. If the quadratic equation has two distinct real roots, which of the following could be true?

I) k=0

II) k=−1

III) k=−3

A) I only
B) III only
C) I and II only
D) II and III only
E) I and III only


Please format the question properly.

Per the question, \(x^2+kx+1=0\) with k=constant.

For a quadratic equation (\(ax^2+bx+c=0\)) to have 2 distinct roots ---> \(D =b^2-4ac > 0\) ---> \(k^2-4>0\) --->\(k<-2\)or \(k>2\) . Only III falls in these ranges. B is the correct answer.

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Re: In the equation x^2+kx+1=0, x is a variable and k is a constant [#permalink]

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New post 14 Oct 2016, 19:08
Engr2012 wrote:
Mascarfi wrote:
In the equation \(x^2+kx+1=0\), x is a variable and k is a constant. If the quadratic equation has two distinct real roots, which of the following could be true?

I) k=0

II) k=−1

III) k=−3

A) I only
B) III only
C) I and II only
D) II and III only
E) I and III only


Please format the question properly.

Per the question, \(x^2+kx+1=0\) with k=constant.

For a quadratic equation (\(ax^2+bx+c=0\)) to have 2 distinct roots ---> \(D =b^2-4ac > 0\) ---> \(k^2-4>0\) --->\(k<-2\)or \(k>2\) . Only III falls in these ranges. B is the correct answer.



how did u get D=b ^ 2-4ac?

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In the equation x^2+kx+1=0, x is a variable and k is a constant [#permalink]

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New post 15 Oct 2016, 01:05
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jonmarrow wrote:
how did u get D=b ^ 2-4ac?


There are some shortcut formulas -

Quote:
1. b2 −4ac < 0 There are no real roots.
2. b2 −4ac = 0 There is one real root.
3. b2 −4ac > 0 There are two real roots.


We are interested only in the third one here...

So, Engr2012 has used it correctly here..

PS : For further query please go through these links -

1. math-algebra-101576.html#p787276
2. http://www.biology.arizona.edu/biomath/ ... roots.html
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Re: In the equation x^2+kx+1=0, x is a variable and k is a constant [#permalink]

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New post 19 Oct 2017, 09:17
Mascarfi wrote:
In the equation \(x^2+kx+1=0\), x is a variable and k is a constant. If the quadratic equation has two distinct real roots, which of the following could be true?

I) k=0

II) k=−1

III) k=−3

A) I only
B) III only
C) I and II only
D) II and III only
E) I and III only


For a quadratic equation in the form of ax^2 + bx + c = 0, the number of real solutions of the equation is determined by the signed value of the quantity b^2 - 4ac, which is known as the discriminant. The property is:

If b^2 - 4ac > 0, then the quadratic equation ax^2 + bx + c = 0 has two distinct real roots.
If b^2 - 4ac = 0, then the quadratic equation ax^2 + bx + c = 0 has one real root.
If b^2 - 4ac < 0, then the quadratic equation ax^2 + bx + c = 0 has no real roots.

Thus, to make sure x^2 + kx + 1 = 0 has two distinct real roots, we have to make sure k^2 - 4(1)(1) is greater than 0 (notice that a = 1, b = k, and c = 1). That is, we want k^2 - 4 > 0. So let’s check the numbers in the given Roman numerals.

I) k = 0: 0^2 - 4 = -4 > 0? (No)

II) k = -1: (-1)^2 - 4 = -3 > 0? (No)

III) k = -3: (-3)^2 - 4 = 5 > 0? (Yes!)

Thus, only Roman numeral III is correct.

Answer: B
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Re: In the equation x^2+kx+1=0, x is a variable and k is a constant   [#permalink] 19 Oct 2017, 09:17
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