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In the equation x^2 + mx + n = 0, x is a variable and m and n are cons

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In the equation x^2 + mx + n = 0, x is a variable and m and n are cons  [#permalink]

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New post 28 Aug 2017, 03:01
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In the equation x^2 + mx + n = 0, x is a variable and m and n are constants. What is the value of n ?

(1) x + 2 is a factor of x^2 + mx + n
(2) The equation x^2 + mx + n = 0 has only one root

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Re: In the equation x^2 + mx + n = 0, x is a variable and m and n are cons  [#permalink]

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New post 28 Aug 2017, 03:16
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1
Bunuel wrote:
In the equation x^2 + mx + n = 0, x is a variable and m and n are constants. What is the value of n ?

(1) x + 2 is a factor of x2 + mx + n
(2) The equation x^2 + mx + n = 0 has only one root



hi..
Bunuel, typo in x2 + mx + n..

lets see the statements

(1) x + 2 is a factor of \(x^2 + mx + n\)
the equation has two roots and 'n' is the product of these TWO roots..
Only one -2 is known
Insuff


(2) The equation \(x^2 + mx + n = 0\) has only one root
this means \(n = root^2\)
insuff


combined

we know that the equation has one root and -2 is a root ..
so n= (-2)^2=4
sufficient
C
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Re: In the equation x^2 + mx + n = 0, x is a variable and m and n are cons  [#permalink]

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New post 28 Aug 2017, 03:21
chetan2u wrote:
Bunuel wrote:
In the equation x^2 + mx + n = 0, x is a variable and m and n are constants. What is the value of n ?

(1) x + 2 is a factor of x2 + mx + n
(2) The equation x^2 + mx + n = 0 has only one root



hi..
Bunuel, typo in x2 + mx + n..

lets see the statements

(1) x + 2 is a factor of \(x^2 + mx + n\)
the equation has two roots and 'n' is the product of these TWO roots..
Only one -2 is known
Insuff


(2) The equation \(x^2 + mx + n = 0\) has only one root
this means \(n = root^2\)
insuff


combined

we know that the equation has one root and -2 is a root ..
so n= (-2)^2=4
sufficient
C

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Typo edited. Thank you!
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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: In the equation x^2 + mx + n = 0, x is a variable and m and n are cons  [#permalink]

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New post 28 Aug 2017, 03:29
Bunuel wrote:
In the equation x^2 + mx + n = 0, x is a variable and m and n are constants. What is the value of n ?

(1) x + 2 is a factor of x^2 + mx + n
(2) The equation x^2 + mx + n = 0 has only one root


(1) gives us x=-2 follows the equation
4-2m+n=0
n-2m=-4
one equation 2 variables
not sufficient

(2) only one root then m^2-4n=0
m^2 = 4n
one equation two variables
not sufficient

on combining
we have
n-2m=-4 & m^2 = 4n

can be solved to get the values of m & n
4n=16+8m=m^2
m^2-8m-16=0
m=4 and n=4
Also since n = r1*r2
but only one root i.e. -2
n=(-2)^2 = 4
C
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Re: In the equation x^2 + mx + n = 0, x is a variable and m and n are cons  [#permalink]

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New post 28 Aug 2017, 03:35
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Ans is C
see the pic for solution
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Re: In the equation x^2 + mx + n = 0, x is a variable and m and n are cons  [#permalink]

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New post 29 Aug 2017, 05:48
(1) x + 2 is a factor of x^2 + mx + n

implies that x = -2
putting the value we get 4-2m+n = 0
insufficient

(2) The equation x^2 + mx + n = 0 has only one root
only one root implies that the roots are equal
for equal root = -b/2a
insufficient

combining 1 & 2
roots are equal and root is -2
so -b/2a = -2
=> -m/2 = -2
m = 4
and putting that value in equation from 1
4-2(4)+n =0
n=4
C is sufficient.
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Re: In the equation x^2 + mx + n = 0, x is a variable and m and n are cons  [#permalink]

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New post 05 Sep 2017, 18:02
1
Bunuel wrote:
In the equation x^2 + mx + n = 0, x is a variable and m and n are constants. What is the value of n ?

(1) x + 2 is a factor of x^2 + mx + n
(2) The equation x^2 + mx + n = 0 has only one root


We need to determine the value of n.

Statement One Alone:

x + 2 is a factor of x^2 + mx + n.

Let f(x) = x^2 + mx + n. If x + 2 is a factor of x^2 + mx + n, then by the factor theorem, f(-2) = 0. That is, (-2)^2 + m(-2) + n = 0 or 4 - 2m + n = 0. However, since we don’t know the value of m, we can’t determine the value of n. Statement one alone is not sufficient to answer the question.

Statement Two Alone:

The equation x^2 + mx + n = 0 has only one root.

Let r be the root. Then we can express x^2 + mx + n as (x - r)^2:

x^2 + mx + n = (x - r)^2

x^2 + mx + n = x^2 - 2rx + r^2

We see that m = -2r and n = r^2. Since we don’t know the value of m or r, we can’t determine the value of n. Statement two alone is not sufficient to answer the question.

Statements One and Two Together:

Since statement one states that x + 2 is a factor of x^2 + mx + n, x = -2 is a root of the equation x^2 + mx + nx = 0. Furthermore, since statement two states that there is only one root, x = -2 must be a double root. That is, x^2 + mx + n = (x + 2)^2:

x^2 + mx + n = (x + 2)^2

x^2 + mx + n = x^2 + 4x + 4

We see that m = 4 and n = 4.

Answer: C
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Re: In the equation x^2 + mx + n = 0, x is a variable and m and n are cons  [#permalink]

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New post 27 Dec 2017, 05:27
Bunuel wrote:
In the equation x^2 + mx + n = 0, x is a variable and m and n are constants. What is the value of n ?

(1) x + 2 is a factor of x^2 + mx + n
(2) The equation x^2 + mx + n = 0 has only one root


Given \(x^2+mx+n=0\) --- equ (1)

Find 'n'

Let r1 & r2 be the roots of the given Quadratic equation (1)
=> Now for a Standard Quad Equation of the form \(ax^2+bx+c=0\)
=> sum of the roots=\(\frac{-b}{a}\)

=> AND product of the roots=\(\frac{c}{a}\)

=> Therefore for equ (1) with roots r1 & r2
=> r1+r2=-m
=> AND r1*r2=n
so if we know the roots of the given equation we can find 'n'

Statement 1 (x + 2) is a factor of \(x^2 + mx + n\)
=> so (x+2)=0
=> OR x=-2= one of the roots
=> let r1=-2
=> Since r2 not known. Therefore Stat 1 NOT SUFFICIENT

Statement 2 The equation \(x^2 + mx + n = 0\) has only one root
=> so both the roots are equal
=> i.e r1=r2
=> n=\(r1^2\)=\(r2^2\)
=> since VALUE of the root not known. Therefore Stat2 NOT SUFFICIENT

BOTH Stat 1 & 2
=> r1=-2 from stat 1
=> n=\(r1^2\) from stat 2
=> Therefore n=\((-2)^2\)=4

Therefore 'C'

Thanks
Dinesh
Re: In the equation x^2 + mx + n = 0, x is a variable and m and n are cons &nbs [#permalink] 27 Dec 2017, 05:27
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