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# One of the roots of the equation x^2 + kx - 6

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One of the roots of the equation x^2 + kx - 6 [#permalink]  23 Jan 2017, 09:05
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One of the roots of the equation $$x^2$$+ $$kx$$ $$- 6 = 0$$ is 3, and k is a constant.

 Quantity A Quantity B The value of k -1

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

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Re: One of the roots of the equation x^2 + kx - 6 [#permalink]  24 Jan 2017, 11:01
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Carcass wrote:

One of the roots of the equation $$x^2$$+ $$kx$$ - 6 = 0 is 3, and k is a constant.

 Quantity A Quantity B The value of k -1

If 3 is a root (solution) to the equation, then x = 3 SATISFIES the equation.
In other words, we know that 3²+ k(3) - 6 = 0
Evaluate to get: 9 + 3k - 6 = 0
Simplify: 3k + 3 = 0
Subtract 3 from both sides: 3k = -3
Divide both sides by 3 to get: k = -1

So, we have:
Quantity A: -1
Quantity B: -1

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Re: One of the roots of the equation x^2 + kx - 6 [#permalink]  10 Nov 2018, 01:21
substitute value of x as 3 and solve for k, K= -1.
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Re: One of the roots of the equation x^2 + kx - 6 [#permalink]  13 Feb 2019, 07:24
just for my understanding, can k=1, as 2 squared +2*1 -6=0? Somebody pls help .
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Re: One of the roots of the equation x^2 + kx - 6 [#permalink]  14 Feb 2019, 03:26
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Could you please elaborate better what do you mean.

It is not that clear.

Regards
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Re: One of the roots of the equation x^2 + kx - 6 [#permalink]  20 Feb 2019, 22:38
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JelalHossain wrote:
just for my understanding, can k=1, as 2 squared +2*1 -6=0? Somebody pls help .

No, k cannot equal 1, because we are told that k is a constant, meaning it's value does not change.

We know that 3 is a root, so when x = 3, the equation works. When x = 3, then k = -1. This means that k is -1, period. It cannot be anything different.

Your equation requires x = 2 and k = 1, but we know this is wrong, because k is not 1, it is -1. We can change the value of x, but not of k.
Re: One of the roots of the equation x^2 + kx - 6   [#permalink] 20 Feb 2019, 22:38
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