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28 May 2007, 22:16
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If x^4 - x^3 -34x^2 + 133x -147<0 , then which one of the following best represents the range for the values of x ?
1)x<3
2)-7<x<3
3)-7/2 <x<3/2
4)-7<x
dono OA.
1)x<3
2)-7<x<3
3)-7/2 <x<3/2
4)-7<x
dono OA.
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Thank you for understanding, and happy exploring!
It's completly out of GMAT scope 
... But, I can try
(B) for me
Let f(x) = x^4 - x^3 -34x^2 + 133x -147.
We need to factorize this terrible expression
. In other words, we need to determine the roots of it.
Now and as it's not 147 but -147 (evident solution would be -1), we can try the limits given in the answer choices. They should give us the value for the roots.
o f(3) = 3^4 - 3^3 - 34*3^2 + 133*3 -147
= 3^3 * (2) - 34*3^2 + 399 - 147
= 3^2 * (6-34) + 399 - 147
= -9 * 28 + 252
= -252 + 252 = 0
Ok, so we have:
o f(x) = (x-3)*(x^3 +b*x^2 +c*x +147/3)
= (x-3)*(x^3 +b*x^2 +c*x +49)
= x^4 + b*x^3 + c*x^2 + 49*x - 3*x^3 - 3*b*x^2 -3*c*x -147
= x^4 + (b-3)*x^3 + (c-3*b)*x^2 + (-3*c+49)*x -147
Thus,
o b-3 = -1
<=> b = 2
and
o c-3*b = -34
<=> c = -34 + 6 = -28
we have,
o f(x) = (x-3)*(x^3 +2x^2 -28*x +49)
We have to continue to factorize... We can try again the given limit.
o (-7)^3 +2(-7)^2 -28*(-7) +49
= 49*(-7+2) + 196 + 49
= 49*(-5+1) + 196
= -196 + 196
= 0
Bingo. We have now:
o f(x) = (x-3)*(x+7)*(x^2 +d*x+7)
= (x-3)*(x^3+d*x^2+7*x +7*x^2+7*d*x +49)
= (x-3)*(x^3 +(d+7)*x^2 + 7*(d+1)*x + 49)
We arrive to:
o d+7 = 2
<=> d = -5
Hence,
o f(x) = (x-3)*(x+7)*(x^2 -5*x+7)
What about x^2 -5*x+7 = 0 ?
o Delta = 25 - 4*7 = -3
Ouf... It means that x^2 -5*x+7 is always from the sign of a=1. Thus,
x^2 -5*x+7 > 0 no matter the value of x
That's also a good news for the sign of f(x):
o sign(f(x)) = sign((x-3)*(x+7))
Finally,
f(x) < 0 when x is between the root -7 and 3.
-7 < x < 3.
... But, I can try
(B) for me
Let f(x) = x^4 - x^3 -34x^2 + 133x -147.
We need to factorize this terrible expression
. In other words, we need to determine the roots of it.
Now and as it's not 147 but -147 (evident solution would be -1), we can try the limits given in the answer choices. They should give us the value for the roots.
o f(3) = 3^4 - 3^3 - 34*3^2 + 133*3 -147
= 3^3 * (2) - 34*3^2 + 399 - 147
= 3^2 * (6-34) + 399 - 147
= -9 * 28 + 252
= -252 + 252 = 0
Ok, so we have:
o f(x) = (x-3)*(x^3 +b*x^2 +c*x +147/3)
= (x-3)*(x^3 +b*x^2 +c*x +49)
= x^4 + b*x^3 + c*x^2 + 49*x - 3*x^3 - 3*b*x^2 -3*c*x -147
= x^4 + (b-3)*x^3 + (c-3*b)*x^2 + (-3*c+49)*x -147
Thus,
o b-3 = -1
<=> b = 2
and
o c-3*b = -34
<=> c = -34 + 6 = -28
we have,
o f(x) = (x-3)*(x^3 +2x^2 -28*x +49)
We have to continue to factorize
... We can try again the given limit.
o (-7)^3 +2(-7)^2 -28*(-7) +49
= 49*(-7+2) + 196 + 49
= 49*(-5+1) + 196
= -196 + 196
= 0
Bingo. We have now:
o f(x) = (x-3)*(x+7)*(x^2 +d*x+7)
= (x-3)*(x^3+d*x^2+7*x +7*x^2+7*d*x +49)
= (x-3)*(x^3 +(d+7)*x^2 + 7*(d+1)*x + 49)
We arrive to:
o d+7 = 2
<=> d = -5
Hence,
o f(x) = (x-3)*(x+7)*(x^2 -5*x+7)
What about x^2 -5*x+7 = 0 ?
o Delta = 25 - 4*7 = -3
Ouf
... It means that x^2 -5*x+7 is always from the sign of a=1. Thus,
x^2 -5*x+7 > 0 no matter the value of x
That's also a good news for the sign of f(x):
o sign(f(x)) = sign((x-3)*(x+7))
Finally,
f(x) < 0 when x is between the root -7 and 3.
-7 < x < 3.
29 May 2007, 01:19
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B!
Best method is backsolving.
Because we have 3 an -7 in the answers try to plot them in the F(x)
if F(3)=0 , f(-7)= 0 then 3 and -7 is FACTORS of the given equation.
Then it looks like as following:
F(x)=(x+3)*(x+7)*(x^2-5x+7)<0
so B.
Best method is backsolving.
Because we have 3 an -7 in the answers try to plot them in the F(x)
if F(3)=0 , f(-7)= 0 then 3 and -7 is FACTORS of the given equation.
Then it looks like as following:
F(x)=(x+3)*(x+7)*(x^2-5x+7)<0
so B.
