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Titleist
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If function x is defined as If F(x) = x^2-(sqrt. 2)x+4 and Updated on: 30 Dec 2004, 19:33
Originally posted by Titleist on 30 Dec 2004, 01:02.
Last edited by Titleist on 30 Dec 2004, 19:33, edited 1 time in total.
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If function x is defined as If F(x) = x^2-(sqrt. 2)x+4 and if F(a) = 0 then is a>2?

1) a is postive

2) a^2 is an integer



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If function x is defined as If F(x) = x^2-(sqrt. 2)x+4 and 30 Dec 2004, 18:23
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Titleist
If function x is defined as If F(x) = x^2-(radical 2)x+4 and if F(a) = 0 then is a>2?

1) a is postive

2) a^2 is an integer
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what is radical 2? is it square root?
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If function x is defined as If F(x) = x^2-(sqrt. 2)x+4 and 30 Dec 2004, 19:34
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MA
Titleist
If function x is defined as If F(x) = x^2-(radical 2)x+4 and if F(a) = 0 then is a>2?

1) a is postive

2) a^2 is an integer

what is radical 2? is it square root?
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yes! :-D
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If function x is defined as If F(x) = x^2-(sqrt. 2)x+4 and 31 Dec 2004, 19:12
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fresinha12
statement 1 is enough..
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please explain for everyone else
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If function x is defined as If F(x) = x^2-(sqrt. 2)x+4 and 31 Dec 2004, 19:38
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Titleist, are you sure there is no typo in the question? To me, x^2-(sqrt. 2)x+4 = 0 has no real solution with the quadratic equation. By the way, I read the above as x^2 - sqrt(2)*x + 4 = 0. If it is so, then there is no real solution. Just think of it as the higher the second term will be, the even more the first term will be and the equation will never be = 0
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If function x is defined as If F(x) = x^2-(sqrt. 2)x+4 and 31 Dec 2004, 19:41
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fresinha12
statement 1 is enough..

please explain for everyone else
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yes it is A.

Given that F(x) = x^2-(sqrt. 2)x+4

then F(a) = a^2-(sqrt. 2) a+4
0 = a^2-(sqrt. 2) a+4
a^2 = (sqrt. 2) a+4
squaring both sides,

a^4 = a+4

from (i)
lets suppose the value of a = 2
a^4 = a+4
16 = 2 + 4
16 = 6 which is not the solution of the equation. therefore, to satisfy the equation a must be less than 2 but positive.

from (ii), the value of a can be both positive and negative.

therefore, the answer is A.




1) a is postive

2) a^2 is an integer
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If function x is defined as If F(x) = x^2-(sqrt. 2)x+4 and Updated on: 31 Dec 2004, 20:55
Originally posted by Paul on 31 Dec 2004, 20:09.
Last edited by Paul on 31 Dec 2004, 20:55, edited 1 time in total.
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MA, your answer has 2 flaws.
1-
Quote:
0 = a^2-(sqrt. 2) a+4
a^2 = (sqrt. 2) a+4
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When you brought a^2 to the left side, you omitted the negative sign of (sqrt. 2)a. This is crucial because if you are squaring both sides, you are looking for another degree answer which omits the fact that there is a root which is negative. You just cannot do this
2-
Quote:
squaring both sides,

a^4 = a+4
Show more

Squaring both sides would have given a^4 = 2a^2 + 4. Alright, this is a minor error only but still, if you take into consideration my first reasoning, this cannot even be.

I believe this question has some problems to it.
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If function x is defined as If F(x) = x^2-(sqrt. 2)x+4 and 31 Dec 2004, 20:22
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quadratic equation is: [b +/- sqrt(b^2 - 4ac)] / 2a
a = 1
b = -sqrt2
c = 4
Just plug in the values and you will see that (b^2 - 4ac) will always have a negative value and you cannot have the sqrt of a negative number.
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If function x is defined as If F(x) = x^2-(sqrt. 2)x+4 and 02 Jan 2005, 15:03
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Paul
MA, your answer has 2 flaws.
1-
Quote:
0 = a^2-(sqrt. 2) a+4
a^2 = (sqrt. 2) a+4
When you brought a^2 to the left side, you omitted the negative sign of (sqrt. 2)a. This is crucial because if you are squaring both sides, you are looking for another degree answer which omits the fact that there is a root which is negative. You just cannot do this
2-
Quote:
squaring both sides,

a^4 = a+4
Squaring both sides would have given a^4 = 2a^2 + 4. Alright, this is a minor error only but still, if you take into consideration my first reasoning, this cannot even be.

I believe this question has some problems to it.
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Paul,

My understanding to the question is as under:

(sqrt. 2) = only square root under (a+4).

with this supposition, squaring both sider of the equation yields,
a^4 = a+4



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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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