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If 4 is one solution of the equation x2 + 3x + k = 10, where
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18 Sep 2012, 02:57
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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
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Re: If 4 is one solution of the equation x2 + 3x + k = 10, where
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18 Sep 2012, 02:57
SOLUTIONIf 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: \(x^2 + 3x +(k10) =0\) 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
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18 Sep 2012, 06:25
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 => (x4)(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
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18 Sep 2012, 03:13
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
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18 Sep 2012, 05:07
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 = 47 = 3 (confirm) Thus Answer A
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Re: If 4 is one solution of the equation x2 + 3x + k = 10, where
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18 Sep 2012, 10:28
Bunuel wrote: The Official Guide for GMAT® Review, 13th Edition  Quantitative Questions ProjectIf 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 GMAT Club is introducing a new project: The Official Guide for GMAT® Review, 13th Edition  Quantitative Questions ProjectEach week we'll be posting several questions from The Official Guide for GMAT® Review, 13th Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution. We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation. Thank you! 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 (X4)(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
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01 Mar 2015, 19:52
You don't need to solve for K, we know that 4 is a solution so we already know one of the terms (x4)(x+......) we also have x2 + 3x + k = 10, 4 + 7=3 so (x4)(x+7):0 hence x: 7



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Re: If 4 is one solution of the equation x2 + 3x + k = 10, where
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18 Apr 2016, 12:48
solving for 4 to find k 16+12+k=10 k=18 (x^2)+3x18=10 testing values you find 7
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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, 07:28
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)(x4) = 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
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27 Jan 2018, 07:11
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
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27 Jan 2018, 08:09
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 = 18So, 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 Cheers, Brent
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Re: If 4 is one solution of the equation x2 + 3x + k = 10, where
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27 Jan 2018, 08:46
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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If 4 is one solution of the equation x2 + 3x + k = 10, where
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Updated on: 26 Feb 2018, 05:38
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: \(x^2 + 3x +(k10) =0\) 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 or your mind blowing explanation niks18
Originally posted by dave13 on 21 Feb 2018, 10:15.
Last edited by dave13 on 26 Feb 2018, 05:38, edited 1 time in total.



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Re: If 4 is one solution of the equation x2 + 3x + k = 10, where
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24 Feb 2018, 07:57
sum of roots is r1+r2= b/a here one root is 4 4+r2= 3/1 r2= 34 = 7



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Re: If 4 is one solution of the equation x2 + 3x + k = 10, where
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26 Feb 2018, 06:21
dave13 wrote: 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: \(x^2 + 3x +(k10) =0\) 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 or your mind blowing explanation niks18 Hi dave13, Here is the generic quadratic equation \(ax^2+bx+c\) and here is your equation \(x^2+16x+64 =0\). Now match both the coefficients of both the equations and tell me what will be the value of a, b & c as per your equation? if you get the values of a, b & c then calculate sum of roots \(= \frac{b}{a}\) and product of roots \(= \frac{c}{a}\)



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If 4 is one solution of the equation x2 + 3x + k = 10, where
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26 Feb 2018, 06:46
niks18 wrote: dave13 wrote: 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: \(x^2 + 3x +(k10) =0\) 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 or your mind blowing explanation niks18 Hi dave13, Here is the generic quadratic equation \(ax^2+bx+c\) and here is your equation \(x^2+16x+64 =0\). Now match both the coefficients of both the equations and tell me what will be the value of a, b & c as per your equation? if you get the values of a, b & c then calculate sum of roots \(= \frac{b}{a}\) and product of roots \(= \frac{c}{a}\) Hi niks18, glad to hear from you \(x^2+16x+64 =0\) from this I can clearly see that \(x^2\) is \(a\) it can be considered as 1 since no number is given 16 is \(b\) 64 is \(c\) so you say > then calculate sum of roots \(= \frac{b}{a}\) ok that means \(\frac{16}{1}\) and product of roots \(= \frac{c}{a}\) and this means \(\frac{64}{1}\) so what next



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If 4 is one solution of the equation x2 + 3x + k = 10, where
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26 Feb 2018, 08:29
Quote: Hi niks18, glad to hear from you \(x^2+16x+64 =0\) from this I can clearly see that \(x^2\) is \(a\) it can be considered as 1 since no number is given 16 is \(b\) 64 is \(c\) so you say > then calculate sum of roots \(= \frac{b}{a}\) ok that means \(\frac{16}{1}\) and product of roots \(= \frac{c}{a}\) and this means \(\frac{64}{1}\) so what next Hi dave13, what is your objective here? Do you want to find out the roots of the equation?. If yes then that could be done by two ways  one by factorization and second by using the discriminant formula. what you did above was you found out the relationship between the roots i.e their sum and product. Now if you are given one root, then using this relationship you can find out the other. So kindly clarify what is your objective



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