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Solving Inequalities related questions can be tricky. In this session, we'll look at different types of Inequalities question that we can solve using number line method.
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The average of (x-1)^2 and (x+3)^2 or 3
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02 Jun 2020, 02:30
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Quantity A
Quantity B
The average of \((x-1)^2\) and \((x+3)^2\)
3
A) Quantity A is greater. B) Quantity B is greater. C) The two quantities are equal. D) The relationship cannot be determined from the information given.
The average of (x-1)^2 and (x+3)^2 or 3
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02 Jun 2020, 05:06
5
Expert Reply
Carcass wrote:
Quantity A
Quantity B
The average of \((x-1)^2\) and \((x+3)^2\)
3
A) Quantity A is greater. B) Quantity B is greater. C) The two quantities are equal. D) The relationship cannot be determined from the information given.
Tricky!!!
Start with: Average \(= \frac{(x-1)^2 + (x+3)^2}{2}\)
Simplify numerator: Average \(= \frac{2x^2 + 4x + 10}{2}\)
Simplify numerator: Average \(= x^2 + 2x + 5\)
This is where it gets tricky!!! You may notice that we kind of have a perfect square hiding in our algebraic expression. We know that \(= x^2 + 2x + 1\) is a perfect square because we can rewrite it as \((x + 1)^2\)
So let's take: \(= x^2 + 2x + 5\) And rewrite it as: \(= x^2 + 2x + 1 + 4\) Now factor the first part to get: \(= (x + 1)^2 + 4\)
So, the average \(= (x + 1)^2 + 4\)
Since \((x + 1)^2 ≥ 0\) for all values of x, we can conclude that \((x + 1)^2 + 4 ≥ 4\) for all values of x
In other words: QUANTITY A: Some number greater than or equal to 4 QUANTITY B: 3
Re: The average of (x-1)^2 and (x+3)^2 or 3
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25 Sep 2021, 05:17
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