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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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Updated on: 10 Aug 2015, 18:03
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55% (hard)

Question Stats:

60% (01:38) correct 40% (02:00) wrong based on 176 sessions

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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

Originally posted by Mascarfi on 10 Aug 2015, 17:53.
Last edited by ENGRTOMBA2018 on 10 Aug 2015, 18:03, edited 1 time in total.
Formatted the question
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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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10 Aug 2015, 18:05
1
2
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

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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14 Oct 2016, 20:08
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

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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15 Oct 2016, 02:05
1
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..

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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19 Oct 2017, 10:17
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.

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Re: In the equation x^2+kx+1=0, x is a variable and k is a constant &nbs [#permalink] 19 Oct 2017, 10:17
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