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If 4 is one solution of the equation x2 + 3x + k = 10, where

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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

Practice Questions
Question: 45
Page: 158
Difficulty: 600
[Reveal] Spoiler: OA

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

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SOLUTION

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

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Slove equation by putting x=4 as it is one of the solution.
u gets value of k=-18

now put k=-18 in the equation,u get eqn in the form of ax^2+bx+c=0. Solving it we get x=4 and x=-7

so other solution is -7.
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Re: If 4 is one solution of the equation x2 + 3x + k = 10, where [#permalink]

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Shortcut for the question-

The sum of roots of a equation is = -b/a = -3
One of the given roots is 4
Thus the other root has to be -7
Check = 4-7 = -3 (confirm)

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

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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.

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

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Bunuel wrote:
The Official Guide for GMAT® Review, 13th Edition - Quantitative Questions Project

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


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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.

other solution is X = -7.
ANswer "A"
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Re: If 4 is one solution of the equation x2 + 3x + k = 10, where [#permalink]

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

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

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New post 27 Jan 2018, 06:11
put 4 in equation and solve for k
So that gives k = -18
put it back in equation and solve for roots. one will come as 4 and another as -7.
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Re: If 4 is one solution of the equation x2 + 3x + k = 10, where [#permalink]

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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


Let's first determine the value of k.

Since x = 4 is a solution to the equation x² + 3x + k = 10, we know that x = 4 SATISFIES the equation.
That is: 4² + 3(4) + k = 10
Evaluate to get: 16 + 12 + k = 10
Solve for k to get: k = -18

So, the ORIGINAL equation is x² + 3x + (-18) = 10
This is the same as: x² + 3x - 18 = 10
We now need to solve this equation.

First, set it equal to zero: x² + 3x - 28 = 0
Factor: (x + 7)(x - 4) = 0
So, x = -7 or x = 4

We already know that x = 4 is one solution.
So, the other solution is x = -7

Answer: A

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

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New post 27 Jan 2018, 07:46
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

\(k = 10 - ( x^2 + 3x )\)

Now, \(k = 10 - ( 16 + 12 )\)

Or, \(k = -18\)

Put the value of k as -18 in the equation and solve now

\(x^2 + 3x - 18 = 10\)

\(x^2 + 3x - 28 = 0\)

Actual solving the equation is not required, we already have one value as 4 , so the other value must be -7 ( As -28 = 4*-7), Answer must be (A)
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Re: If 4 is one solution of the equation x2 + 3x + k = 10, where [#permalink]

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New post 21 Feb 2018, 09:15
Bunuel wrote:
SOLUTION

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.


Hi Bunuel :) , i understand how you soled it, but regardimg Vieta`s properties \(x_1+x_2=-\frac{b}{a}\) AND \(x_1*x_2=\frac{c}{a}\)

how can i use these properties to find roots for example of this equation \(x^2+16x+64 =0\)

if i follow the first property \(x_1+x_2=-\frac{b}{a}\) i get \(x_1+x_2 = \frac{64}{1}\) the same question to the second property ...


on the other hand i know tis rule ---> w

So 64 is our C term
The numbers that can multiply to make 64 are +8 and +8 (8*8 = 64)

16 is our B term

Now find the two factors of C that add up to your B term 8 and 8 Hence 8+8 = 16


Now plug in values I have chosen into factored equation (x+a) (x+b)= 0

(x+8) (x+8)= 0
Now solve for X by equating to 0
(x+8) =0 ---- > x = -8
(x+8)= 0 ----- >x = - 8


But i dont understand how to use these properties to get the same values( -8; -8) \(x_1+x_2=-\frac{b}{a}\) AND \(x_1*x_2=\frac{c}{a}\)

I would appreciate your fantastic explanation :)
Re: If 4 is one solution of the equation x2 + 3x + k = 10, where   [#permalink] 21 Feb 2018, 09:15
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