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The ans is 5

putting the values gives us 2m^2-10m-3=0

solving this gives us m= (10+sqrt(124))/4 and (10-sqrt(124))/4

21/4-1/4 =5 ans
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Bunuel

If \(f(x) = 5x^2 - 4\)and \(g(x) = 3x + 1\), then what is the sum of all the values of \(m\) for which \(f(m - 1) = g(m^2 + 1)\) ?

A. -10
B. -5
C. 0
D. 5
E. 10

Kudos for a correct solution.


f(m-1)= 5(m-1)^2 -4
= 5(m^2+1-2m) -4
=5m^2+5-10m-4
= 5m^2-10m +1

g(m^2+1)= 3(m^2+1) +1
=3m^2+4

now f(m-1)=g(m^2+1)
5m^2-10m +1 = 3m^2+4
2m^2-10m -3=0

also, any quadratic equation can be written as
x^2-(sum of the roots)x + product of the roots

here since coeff. of m^2 is 2, therefore we will divide the whole expression by 2
and we will get, m^2-5m-3/2=0

thus sum of roots = 5

hence D
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Substituting (m-1) and (m^2+1) in place of x in the functions f(x) and g(x) respectively,

5(m-1)^2-4=3(m^2+1)
5m^2-10m+5-4=3m^2+3+1
2m^2-10m=0
2m(m-5)=0

2m=0 and m-5=0

The roots of m are 0 and 5 and their sum is 5

Answer D
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Subbing (m-1) into f(x) yields 5m^2 - 10m + 1

Subbing (m^2 + 1) into g(x) yields 3m^2 + 4

Setting f(m-1) equal to g(m^2 + 1) yields
3m^2 + 1 = 5m^2 - 10m + 1

This simplifies to
0 = 2m^2 - 10m - 3

Using Viete's rules, we know that the sum of the roots x1 and x2 in ax^2 + bx + c is equal to -b/a, hence
x1 + x2 = -(-10)/2 = 5

This is the desired answer. Therefore, D.
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\(f(x) = 5x^2 - 4\)

\(f(m-1) = 5 (m-1)^2 - 4\)

\(= 5(m^2 - 2m + 1) - 4\)

\(= 5m^2 - 10m + 5 - 4\)

\(= 5m^2 - 10m + 1\) .............. (1)

g(x) = 3x+1

\(g(m^2 + 1) = 3(m^2 + 1) + 1\)

\(= 3m^2 + 4\) ............ (2)

Equating (1) & (2)

\(5m^2 - 10m + 1 = 3m^2 + 4\)

\(2m^2 - 10m - 3 = 0\)

Sum of roots \(= \frac{-(-10)}{2} = 5\)

Answer = D
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fra
Substituting (m-1) and (m^2+1) in place of x in the functions f(x) and g(x) respectively,

5(m-1)^2-4=3(m^2+1)
5m^2-10m+5-4=3m^2+3+1
2m^2-10m=0
2m(m-5)=0

2m=0 and m-5=0

The roots of m are 0 and 5 and their sum is 5

Answer D

Equation would come up as \(2m^2 - 10m - 3\)

Answer = D, however individual roots cannot be 0 & 5
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Thanks for pointing that out!
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If \(f(x) = 5x^2 - 4\) and \(g(x) = 3x + 1\), then what is the sum of all the values of \(m\) for which \(f(m - 1) = g(m^2 + 1)\) ?

A. -10
B. -5
C. 0
D. 5
E. 10

Kudos for a correct solution.

[/quote]

f(m-1) = g(m^2+1)
5(m-1)^2-4=3(m^2+1)+1
2m^2 - 10m - 3 = 0
Sum of roots is = -b/a
so 10/2 = 5
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f(m-1) = g(m^2+1)
5(m-1)^2-4=3(m^2+1)+1
2m^2 - 10m - 3 = 0
For equation aX^2+bX+C=0 sum of roots =(-b/a)

so here sum of roots= -(-10/2) = 5
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Bunuel

If \(f(x) = 5x^2 - 4\) and \(g(x) = 3x + 1\), then what is the sum of all the values of \(m\) for which \(f(m - 1) = g(m^2 + 1)\) ?

A. -10
B. -5
C. 0
D. 5
E. 10

Kudos for a correct solution.


Pleas find attached the solution.

My opinion: it's highly unlikely to see a question based on these concepts in actual GMAT exam.

Posted from my mobile device
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GMATinsight


Pleas find attached the solution.

My opinion: it's highly unlikely to see a question based on these concepts in actual GMAT exam.


Dear GMATInsight,

I'd like to draw you attention that there is an error in f(x) function. You have changed it from (-4) to (+4). The final quadratic equation should be:

2m^2 - 10m - 3 = 0
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Mo2men
GMATinsight


Pleas find attached the solution.

My opinion: it's highly unlikely to see a question based on these concepts in actual GMAT exam.


Dear GMATInsight,

I'd like to draw you attention that there is an error in f(x) function. You have changed it from (-4) to (+4). The final quadratic equation should be:

2m^2 - 10m - 3 = 0

Thanks for notifying my mistake... However the pleasant surprise is that answer still doesn't change as the answer is dependent on the coefficients of x^2 and x...

So I hope that is ok if I accept the mistake made in first step however endorse the method to find answer (roots) of a quadratic equation. :)
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Given \(f\left(x\right)\ =\ 5x^2-4 \) \(g\left(x\right)\ =\ 3x+1\)
Substituting the function for
\(f\left(m-1\right)\ =\ 5\left(m-1\right)^2-4\) = \(5\left(m^2-2m+1\right)-4\)
= \(5m^2-10m+1\)
Similarly for \(g\left(m^2+1\right)\)
= \(3\left(m^2+1\right)+1\)
Equating the two we have
\(2m^2-10m-3\ =\ 0\)
The sum of possible values of m is equal to the sum of roots of the equation which is equal to 5.
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This question falls under what topic please?
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Adesewaa
This question falls under what topic please?


Check the category tags, the question belongs to Algebra and Functions and Custom Characters category:


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