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705-805 (Hard)|   Algebra|      
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Bunuel
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In the equation x^2+bx+c=0, b and c are constants, and x is 04 Mar 2021, 01:15
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95% (hard)

Question Stats:

40% (01:57) correct 60% (01:38) wrong based on 215 sessions

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In the equation \(x^2+bx+c=0\), b and c are constants, and x is a variable. If m and k are the roots of \(x^2+bx+c=0\), then m−k=?

(1) c = −10
(2) b = 3



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In the equation x^2+bx+c=0, b and c are constants, and x is 05 Mar 2021, 01:30
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Bunuel
In the equation \(x^2+bx+c=0\), b and c are constants, and x is a variable. If m and k are the roots of \(x^2+bx+c=0\), then m−k=?

(1) c = −10
(2) b = 3



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Even when we combine the statements we get \(x^2+3x-10=0\) --> \((x+5)(x-2)=0\) --> \(x=-5\) or \(x=2\) --> m and k are -5 and 2, but we don't know which one is which --> m-k is either -5-2=-7 or 2-(-5) = 7.

Answer: E.
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In the equation x^2+bx+c=0, b and c are constants, and x is 04 Mar 2021, 01:34
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Bunuel
In the equation \(x^2+bx+c=0\), b and c are constants, and x is a variable. If m and k are the roots of \(x^2+bx+c=0\), then m−k=?

(1) c = −10
(2) b = 3
Show more

Question: m-k = ?, where m and k are the roots of the equation \(x^2+bx+c=0\)

FOr any Quadratic Equation of the type \(ax^2+bx+c = 0\)


Sum of the roots i.e. \(α+β = \frac{-b}{a}\)
Product of the roots i.e. \(α*β = \frac{c}{a}\)

Also, \((α-β)^2 = (α+β)^2 - 4*αβ\)



So to find m-k, we need Sum of the roots as well as the product of the roots

i.e. we require values of both b and c while a = 1 in \(x^2+bx+c=0\)


Statement (1) c = −10
NOT SUFFICIENT

Statement (2) b = 3
NOT SUFFICIENT

Combining the statements

We have both b and c hence

SUFFICIENT

Answer: Option C
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In the equation x^2+bx+c=0, b and c are constants, and x is 04 Mar 2021, 03:14
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I think this is E. IMO, this was very tricky.

From both statements, we know that the equation is x^2 + 3x -10=0. This means the roots are 2 and -5. However, we don't know which is m and which is k so the answer could be 2-(-5)=7 or -5-2=-7. Hence, E.
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In the equation x^2+bx+c=0, b and c are constants, and x is 04 Mar 2021, 03:15
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Bunuel
In the equation \(x^2+bx+c=0\), b and c are constants, and x is a variable. If m and k are the roots of \(x^2+bx+c=0\), then m−k=?

(1) c = −10
(2) b = 3

Question: m-k = ?, where m and k are the roots of the equation \(x^2+bx+c=0\)

FOr any Quadratic Equation of the type \(ax^2+bx+c = 0\)


Sum of the roots i.e. \(α+β = \frac{-b}{a}\)
Product of the roots i.e. \(α*β = \frac{c}{a}\)

Also, \((α-β)^2 = (α+β)^2 - 4*αβ\)



So to find m-k, we need Sum of the roots as well as the product of the roots

i.e. we require values of both b and c while a = 1 in \(x^2+bx+c=0\)


Statement (1) c = −10
NOT SUFFICIENT

Statement (2) b = 3
NOT SUFFICIENT

Combining the statements

We have both b and c hence

SUFFICIENT

Answer: Option C
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Roots are 5 and -2. If m = 5, k = -2, m-k = 7.

If m = -2, k = 5, then, m-k = -7.

Not sure how C is the answer?
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In the equation x^2+bx+c=0, b and c are constants, and x is 04 Mar 2021, 13:07
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If m and k are the roots of x^2+bx+c=0

m+k = b and m*k= c
(m-k)^2 = (m+k)^ - 4m*k = b^2 -4c

Stat1: c = −10, we need value of b also. Not sufficient.
Stat2: b = 3, we need value of c also. Not sufficient.

Combining both, (m-k)^2 = b^2 -4c = 9+40 = 49 so, (m-k) = +-7. Not sufficient.

So, I think E. :)
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In the equation x^2+bx+c=0, b and c are constants, and x is 12 Mar 2021, 00:06
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Bunuel
In the equation \(x^2+bx+c=0\), b and c are constants, and x is a variable. If m and k are the roots of \(x^2+bx+c=0\), then m−k=?

(1) c = −10
(2) b = 3

Question: m-k = ?, where m and k are the roots of the equation \(x^2+bx+c=0\)

FOr any Quadratic Equation of the type \(ax^2+bx+c = 0\)


Sum of the roots i.e. \(α+β = \frac{-b}{a}\)
Product of the roots i.e. \(α*β = \frac{c}{a}\)

Also, \((α-β)^2 = (α+β)^2 - 4*αβ\)



So to find m-k, we need Sum of the roots as well as the product of the roots

i.e. we require values of both b and c while a = 1 in \(x^2+bx+c=0\)


Statement (1) c = −10
NOT SUFFICIENT

Statement (2) b = 3
NOT SUFFICIENT

Combining the statements

We have both b and c hence

SUFFICIENT

Answer: Option C
Show more


GMATinsight EducationAisle I followed the same logic and landed with (C) but here the O.A is (E). Bunuel stated that we don't know the individual values of m and k and hence (E) is the answer. BUT do we need to know the individual values? I mean.... we just need to find m-k and we certainly can find the value without knowing the actual values of m and k
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In the equation x^2+bx+c=0, b and c are constants, and x is 08 Aug 2023, 07:42
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Bunuel
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In the equation \(x^2+bx+c=0\), b and c are constants, and x is a variable. If m and k are the roots of \(x^2+bx+c=0\), then m−k=?

(1) c = −10
(2) b = 3

Even when we combine the statements we get \(x^2+3x-10=0\) --> \((x+5)(x-2)=0\) --> \(x=-5\) or \(x=2\) --> m and k are -5 and 2, but we don't know which one is which --> m-k is either -5-2=-7 or 2-(-5) = 7.

Answer: E.
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Hi Bunuel

As per (I) C = -10 ==> m*k = -10

m = -1, k = 10 (m-k = -11)
m = -2, k = 5 (m-k = -7)
m = 1, k = -10 (m-k = 11)
m = 2, k = -5 (m-k = 7)

Not Sufficient

As per (II) b = -3 ==> m+k = -3

-1000+997 = -3 (m-k = -1997)
1-4 = -3 (m-k = 5)

Not Sufficient

But When we combine (I) + (II)
m = -1, k = 10 (m+k = 9)
m = -2, k = 5 (m+k = 3)
m = 1, k = -10 (m+k = -9)
m = 2, k = -5 (m+k = -3)

hence m-k = 7

Why is this approach wrong?
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In the equation x^2+bx+c=0, b and c are constants, and x is 12 Aug 2023, 03:40
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SatvikVedala
Bunuel
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In the equation \(x^2+bx+c=0\), b and c are constants, and x is a variable. If m and k are the roots of \(x^2+bx+c=0\), then m−k=?

(1) c = −10
(2) b = 3
Even when we combine the statements we get \(x^2+3x-10=0\) --> \((x+5)(x-2)=0\) --> \(x=-5\) or \(x=2\) --> m and k are -5 and 2, but we don't know which one is which --> m-k is either -5-2=-7 or 2-(-5) = 7.

Answer: E.

Hi Bunuel

As per (I) C = -10 ==> m*k = -10

m = -1, k = 10 (m-k = -11)
m = -2, k = 5 (m-k = -7)
m = 1, k = -10 (m-k = 11)
m = 2, k = -5 (m-k = 7)

Not Sufficient

As per (II) b = -3 ==> m+k = -3

-1000+997 = -3 (m-k = -1997)
1-4 = -3 (m-k = 5)

Not Sufficient

But When we combine (I) + (II)
m = -1, k = 10 (m+k = 9)
m = -2, k = 5 (m+k = 3)
m = 1, k = -10 (m+k = -9)
m = 2, k = -5 (m+k = -3)

hence m-k = 7

Why is this approach wrong?
Show more

m*k = -10 and m + k = -3 has two solutions: k = -5, m = 2 and k = 2, m = -5. Therefore, m - k is either -5 - 2 = -7 or 2 - (-5) = 7.
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In the equation x^2+bx+c=0, b and c are constants, and x is 17 Mar 2025, 08:20
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