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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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Updated on: 18 Sep 2017, 09:25
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Question Stats:

53% (02:14) correct 47% (02:26) wrong based on 226 sessions

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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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18 Sep 2017, 09:32
1
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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18 Sep 2017, 11:11
1
2
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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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.) $$g(x) ≥ 5b^2$$

$$g(x) = x^2 – 4bx + 9b^2$$

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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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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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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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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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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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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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   [#permalink] 08 May 2018, 18:46
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