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If 4 is one solution of the equation x^2 + 3x + k = 10, where k is a constant, what is the other solution?

(A) -7 (B) -4 (C) -3 (D) 1 (E) 6

Viete's formula for the roots \(x_1\) and \(x_2\) of equation \(ax^2+bx+c=0\): \(x_1+x_2=-\frac{b}{a}\) AND \(x_1*x_2=\frac{c}{a}\).

Rewrite the equation given: \(x^2 + 3x +(k-10) =0\) Thus according to the first property \(4+x_2=-\frac{3}{1}\) --> \(x_2=-7\). As we can see we can find the value of \(x_2\) even without finding the value of \(k\).

Re: If 4 is one solution of the equation x2 + 3x + k = 10, where [#permalink]

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18 Sep 2012, 05:25

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Bunuel wrote:

If 4 is one solution of the equation x^2 + 3x + k = 10, where k is a constant, what is the other solution?

(A) -7 (B) -4 (C) -3 (D) 1 (E) 6

Since 4 is one of the solution, so it must satisfy the equation. Putting 4 as value of x: 4^2 + 3*4 + k = 10 => k =-18 so the equations is x^2 + 3x - 28 = 0

Solving this by factorization gives => (x-4)(x+7) = 0 So x = -7 is other solution.

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One sol. of equation is 4. therefor putting this as X in the equation and finout the constant K. therefore K = -28. so, equation can be rewritten as (X-4)(X+7) = 0.

If 4 is one solution of the equation x^2 + 3x + k = 10, where k is a constant, what is the other solution?

(A) -7 (B) -4 (C) -3 (D) 1 (E) 6

Viete's formula for the roots \(x_1\) and \(x_2\) of equation \(ax^2+bx+c=0\): \(x_1+x_2=-\frac{b}{a}\) AND \(x_1*x_2=\frac{c}{a}\).

Rewrite the equation given: \(x^2 + 3x +(k-10) =0\) Thus according to the first property \(4+x_2=-\frac{3}{1}\) --> \(x_2=-7\). As we can see we can find the value of \(x_2\) even without finding the value of \(k\).

Answer: A.

Kudos points given to everyone with correct solution. Let me know if I missed someone.
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Re: If 4 is one solution of the equation x2 + 3x + k = 10, where [#permalink]

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23 Oct 2013, 05:03

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Re: If 4 is one solution of the equation x2 + 3x + k = 10, where [#permalink]

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Hello from the GMAT Club BumpBot!

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Re: If 4 is one solution of the equation x2 + 3x + k = 10, where [#permalink]

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01 Mar 2015, 18:52

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You don't need to solve for K, we know that 4 is a solution so we already know one of the terms (x-4)(x+......) we also have x2 + 3x + k = 10, -4 + 7=3 so (x-4)(x+7):0 hence x: -7

Re: If 4 is one solution of the equation x2 + 3x + k = 10, where [#permalink]

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17 Apr 2016, 21:04

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If 4 is one solution of the equation x2 + 3x + k = 10, where [#permalink]

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18 Apr 2016, 11:48

solving for 4 to find k 16+12+k=10 k=-18 (x^2)+3x-18=10 testing values you find -7
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Re: If 4 is one solution of the equation x2 + 3x + k = 10, where [#permalink]

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28 Apr 2016, 06:28

Bunuel wrote:

If 4 is one solution of the equation x^2 + 3x + k = 10, where k is a constant, what is the other solution?

(A) -7 (B) -4 (C) -3 (D) 1 (E) 6

Practice Questions Question: 45 Page: 158 Difficulty: 600

The phrase “4 is one solution of the equation” means that one value of x is 4. Thus, we first must plug 4 for x into the given equation to determine the value of k. So we have

4^2 + (3)(4) + k = 10

16 + 12 + k = 10

28 + k = 10

k = -18

Next we plug -18 into the given equation for k and then solve for x.

x^2 + 3x – 18 = 10

x^2 + 3x – 28 = 0

(x+7)(x-4) = 0

x = -7 or x = 4

Thus, -7 is the other solution. Answer A.
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Jeffrey Miller Jeffrey Miller Head of GMAT Instruction

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Re: If 4 is one solution of the equation x2 + 3x + k = 10, where
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28 Apr 2016, 06:28

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