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In the equation x^2+bx+c=0, b and c are constants, and x is
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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 [#permalink] New post  04 Mar 2021, 00:15
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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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Re: In the equation x^2+bx+c=0, b and c are constants, and x is [#permalink] New post  05 Mar 2021, 00:30
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Bunuel wrote:
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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Re: In the equation x^2+bx+c=0, b and c are constants, and x is [#permalink] New post  04 Mar 2021, 00:34
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Bunuel wrote:
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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Re: In the equation x^2+bx+c=0, b and c are constants, and x is [#permalink] New post  04 Mar 2021, 02:14
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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Re: In the equation x^2+bx+c=0, b and c are constants, and x is [#permalink] New post  04 Mar 2021, 02:15
GMATinsight wrote:
Bunuel wrote:
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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Re: In the equation x^2+bx+c=0, b and c are constants, and x is [#permalink] New post  04 Mar 2021, 12:07
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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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Re: In the equation x^2+bx+c=0, b and c are constants, and x is [#permalink] New post  11 Mar 2021, 23:06
GMATinsight wrote:
Bunuel wrote:
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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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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Re: In the equation x^2+bx+c=0, b and c are constants, and x is [#permalink]
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