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If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then

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If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then  [#permalink]

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If \(g(x) = x^2 – 4bx + 9b^2\), where \(b\) is any positive integer, then which of the following must be true

A.) \(g(x) < -b\)

B.) \(g(x) < 0\)

C.) \(0 < g(x) < 4b\)

D.) \(g(x) ≥ 4b\)

E.) \(g(x) ≥ 5b^2\)

Originally posted by niks18 on 17 Sep 2017, 08:04.
Last edited by niks18 on 18 Sep 2017, 09:25, edited 1 time in total.
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Re: If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then  [#permalink]

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New post 18 Sep 2017, 09:32
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1
niks18 wrote:
If \(g(x) = x^2 – 4bx + 9b^2\), where \(b\) is any positive integer, then which of the following must be true

A.) \(g(x) < -b\)

B.) \(g(x) < 0\)

C.) \(0 < g(x) < 4b\)

D.) \(g(x) ≥ 4b\)

E.) \(g(x) ≥ 5b^2\)





g(x)= x^2-4bx+9b^2--> (x-2b)^2+5b^2.
Since b is a positive integer, value of g(x) will always be greater than 5b^2 as (x-2b)^2 will alwz be positive.
Hence for min value of b i.e 1, g(x)=5b^2.
Hence E. g(x)>=5b^2
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Re: If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then  [#permalink]

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New post 18 Sep 2017, 11:11
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niks18 wrote:
If \(g(x) = x^2 – 4bx + 9b^2\), where \(b\) is any positive integer, then which of the following must be true

A.) \(g(x) < -b\)

B.) \(g(x) < 0\)

C.) \(0 < g(x) < 4b\)

D.) \(g(x) ≥ 4b\)

E.) \(g(x) ≥ 5b^2\)


OE

\(g(x) = x^2 – 4bx + 4b^2+5b^2\)

or, \(g(x) = (x – 2b)^2+5b^2\),

Now, \((x-2b)^2≥0\), for all values of \(x\). Adding \(5b^2\) to both sides of the inequality we get

\((x-2b)^2+5b^2≥5b^2\)

Or, \(g(x)≥5b^2\)

Option E
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Re: If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then  [#permalink]

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New post 23 Sep 2017, 03:27
niks18 wrote:
If \(g(x) = x^2 – 4bx + 9b^2\), where \(b\) is any positive integer, then which of the following must be true

A.) \(g(x) < -b\)

B.) \(g(x) < 0\)

C.) \(0 < g(x) < 4b\)

D.) \(g(x) ≥ 4b\)

E.) \(g(x) ≥ 5b^2\)


Answer is E as follows.

\(g(x) = x^2 – 4bx + 9b^2\)

Since the above is a quadratic equation, its maxima/minima will be at (-)-4d/2 = 2b. Also since b is a +ve integer 2b will be +ve. Hence as per the rule the quadratic equation will have minimum value at x=2b.
Putting the value of 2b back into the equation will yield g(2b) = \((2b)^2\) - 4b*(2b) + 9\(b^2\) = 5\(b^2\)
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Re: If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then  [#permalink]

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New post 28 Apr 2018, 22:24
+1 for option E. The given expression can be expressed as (x-2b)^2+5b^2. The first part is always positive. The expression then is always greater than 5(b^2). Hence option E it is.
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Re: If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then  [#permalink]

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New post 07 May 2018, 20:02
niks18 wrote:
niks18 wrote:
If \(g(x) = x^2 – 4bx + 9b^2\), where \(b\) is any positive integer, then which of the following must be true

A.) \(g(x) < -b\)

B.) \(g(x) < 0\)

C.) \(0 < g(x) < 4b\)

D.) \(g(x) ≥ 4b\)

E.) \(g(x) ≥ 5b^2\)


OE

\(g(x) = x^2 – 4bx + 4b^2+5b^2\)

or, \(g(x) = (x – 2b)^2+5b^2\),

Now, \((x-2b)^2≥0\), for all values of \(x\). Adding \(5b^2\) to both sides of the inequality we get

\((x-2b)^2+5b^2≥5b^2\)

Or, \(g(x)≥5b^2\)

Option E




Can you please further explain the need to factor the equation? I selected E because 9b^2 is the original equation. Any help would be super appreciated.

Thanks

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Re: If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then  [#permalink]

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New post 07 May 2018, 22:19
syedazeem3 wrote:
niks18 wrote:
niks18 wrote:
If \(g(x) = x^2 – 4bx + 9b^2\), where \(b\) is any positive integer, then which of the following must be true

A.) \(g(x) < -b\)

B.) \(g(x) < 0\)

C.) \(0 < g(x) < 4b\)

D.) \(g(x) ≥ 4b\)

E.) \(g(x) ≥ 5b^2\)


OE

\(g(x) = x^2 – 4bx + 4b^2+5b^2\)

or, \(g(x) = (x – 2b)^2+5b^2\),

Now, \((x-2b)^2≥0\), for all values of \(x\). Adding \(5b^2\) to both sides of the inequality we get

\((x-2b)^2+5b^2≥5b^2\)

Or, \(g(x)≥5b^2\)

Option E




Can you please further explain the need to factor the equation? I selected E because 9b^2 is the original equation. Any help would be super appreciated.

Thanks

Posted from my mobile device


hi syedazeem3

will be glad to clarify your doubts. I would like to understand how you deduced option E simply by looking at \(9b^2\) in the original equation. If instead of \(5b^2\), option E had mentioned \(4b^2\), then would you still have chosen option E as correct?
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Re: If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then  [#permalink]

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New post 08 May 2018, 05:05
niks18 wrote:
syedazeem3 wrote:
niks18 wrote:
[quote="niks18"]If \(g(x) = x^2 – 4bx + 9b^2\), where \(b\) is any positive integer, then which of the following must be true

A.) \(g(x) < -b\)

B.) \(g(x) < 0\)

C.) \(0 < g(x) < 4b\)

D.) \(g(x) ≥ 4b\)

E.) \(g(x) ≥ 5b^2\)


OE

\(g(x) = x^2 – 4bx + 4b^2+5b^2\)

or, \(g(x) = (x – 2b)^2+5b^2\),

Now, \((x-2b)^2≥0\), for all values of \(x\). Adding \(5b^2\) to both sides of the inequality we get

\((x-2b)^2+5b^2≥5b^2\)

Or, \(g(x)≥5b^2\)

Option E




Can you please further explain the need to factor the equation? I selected E because 9b^2 is the original equation. Any help would be super appreciated.

Thanks

Posted from my mobile device


hi syedazeem3

will be glad to clarify your doubts. I would like to understand how you deduced option E simply by looking at \(9b^2\) in the original equation. If instead of \(5b^2\), option E had mentioned \(4b^2\), then would you still have chosen option E as correct?[/quote]


My logic was, regardless of what is before 9b^2, 5b^2 would be smaller - the original equation would have to be larger than 5b^2. I would’ve probably picked E if it was 4b^2.


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Re: If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then  [#permalink]

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New post 08 May 2018, 07:03
1
niks18 wrote:
If \(g(x) = x^2 – 4bx + 9b^2\), where \(b\) is any positive integer, then which of the following must be true

A.) \(g(x) < -b\)

B.) \(g(x) < 0\)

C.) \(0 < g(x) < 4b\)

D.) \(g(x) ≥ 4b\)

E.) \(g(x) ≥ 5b^2\)

OE

\(g(x) = x^2 – 4bx + 4b^2+5b^2\)

or, \(g(x) = (x – 2b)^2+5b^2\),

Now, \((x-2b)^2≥0\), for all values of \(x\). Adding \(5b^2\) to both sides of the inequality we get

\((x-2b)^2+5b^2≥5b^2\)

Or, \(g(x)≥5b^2\)

Option E


syedazeem3 wrote:
My logic was, regardless of what is before 9b^2, 5b^2 would be smaller - the original equation would have to be larger than 5b^2. I would’ve probably picked E if it was 4b^2.


Hi syedazeem3

Note this is an algebraic equation where \(x\) is a variable and \(b\) is any constant and the question asks which relation for the equation ALWAYS HOLDS TRUE.

By your logic if Option E had been \(8b^2\), then you would have selected it which is NOT ALWAYS TRUE

Suppose \(x=2\) & \(b=1\) then

\(8b^2=8*1^2=8\), but

\(g(x) = x^2 – 4bx + 9b^2=2^2-4*2*1+9*1^2=5\)

clearly here \(g(x)<8b^2\) in this case; but \(g(x)=5b^2\)

So until you factories this quadratic equation you will not come to know that the function \(g(x)\) is always greater than or equal to \(5b^2\) for any value of \(x\) & \(b\)
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Re: If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then  [#permalink]

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New post 08 May 2018, 18:46
niks18 wrote:
niks18 wrote:
If \(g(x) = x^2 – 4bx + 9b^2\), where \(b\) is any positive integer, then which of the following must be true

A.) \(g(x) < -b\)

B.) \(g(x) < 0\)

C.) \(0 < g(x) < 4b\)

D.) \(g(x) ≥ 4b\)

E.) \(g(x) ≥ 5b^2\)

OE

\(g(x) = x^2 – 4bx + 4b^2+5b^2\)

or, \(g(x) = (x – 2b)^2+5b^2\),

Now, \((x-2b)^2≥0\), for all values of \(x\). Adding \(5b^2\) to both sides of the inequality we get

\((x-2b)^2+5b^2≥5b^2\)

Or, \(g(x)≥5b^2\)

Option E


syedazeem3 wrote:
My logic was, regardless of what is before 9b^2, 5b^2 would be smaller - the original equation would have to be larger than 5b^2. I would’ve probably picked E if it was 4b^2.


Hi syedazeem3

Note this is an algebraic equation where \(x\) is a variable and \(b\) is any constant and the question asks which relation for the equation ALWAYS HOLDS TRUE.

By your logic if Option E had been \(8b^2\), then you would have selected it which is NOT ALWAYS TRUE

Suppose \(x=2\) & \(b=1\) then

\(8b^2=8*1^2=8\), but

\(g(x) = x^2 – 4bx + 9b^2=2^2-4*2*1+9*1^2=5\)

clearly here \(g(x)<8b^2\) in this case; but \(g(x)=5b^2\)

So until you factories this quadratic equation you will not come to know that the function \(g(x)\) is always greater than or equal to \(5b^2\) for any value of \(x\) & \(b\)




Thanks for the help here niks18. Appreciate the insight - finally understood this.

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Re: If g(x)= x^2 – 4bx + 9b^2, where b is any positive integer, then &nbs [#permalink] 08 May 2018, 18:46
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