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secondary 3 | A Maths
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jq
Jq

secondary 3 chevron_right A Maths chevron_right Singapore

pls help tia

Date Posted: 2 years ago
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Eric Nicholas K
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jq
Jq
2 years ago
hi thank you for your help !! i dont quite understand the greater or equal to 12 and below that part
Eric Nicholas K
Eric Nicholas K
2 years ago
Give me a moment while I type this. It will take quite some time.
Eric Nicholas K
Eric Nicholas K
2 years ago
Whenever we have squared expressions such as x², (x + 1)², (x + 2)², (x - 1)², (x - 2)², (2x + 1)² and so on, we will always have non-negative output values. This is a useful property of squared functions which you must know.

Consider x². What happens when we input different values of x?

Try taking 1². Its value is 1.
Try taking 2². Its value is 4.
Try taking 0². Its value is 0.
Try taking (-1)². Its value is 1.

[Note that -1² = -1 is another thing altogether and it's not the same scenario]

Try taking (-2)². Its value is 4.

Think of whether it would be possible for us to obtain a negative output value.

Howsoever you try this, you will never get a negative number. This is because when we square a negative number, we are essentially multiplying two negative numbers - and the negative signages cancel each other every time.

So, the lowest possible value of a squared expression is 0.

However, for this function, it's not that simple.

More to come...
Eric Nicholas K
Eric Nicholas K
2 years ago
Now, our discriminant of roots is p² - 6p + 21. We need to be careful here.

Just because p² has a lowest possible value of 0 (and therefore p² + 21 has a lowest possible value of 21) does not mean that the lowest possible value of p² - 6p + 21 will be 21. The culprit is the 6p term.

(Of course, if the lowest possible value of p² is 0, then adding 12 to this p² gives us the lowest possible value of p² + 21, which is 0 + 21 = 21]

We need to find a way to get rid of the 6p somehow whilst maintaining mathematical equality in each statement we make. This is done by completing the square.

[Another method is to find THE DISCRIMINANT of this discriminant, which I will not cover here]

p² - 6p + 21
= p² - 6p + 3² - 3² + 21

[we must ensure that the coefficient of p is 1, or at least a perfect square; then we can say that half of 6 is 3, so we introduce 3² to complete the square and return 3² to maintain equality]

= (p - 3)² - 3² + 21

[the point of the introduction of the 3²; it will get absorbed into the factorisation]

= (p - 3)² + 12

Now we got rid of the 6p successfully. We see that (p - 3)², being a squared expression, has a lowest possible value of 0. This is achieved when p = 3. Try other values of p; you will never be able to get a negative value out of the term (p - 3)².

So, the lowest possible value of (p - 3)² is 0.
Add 12 to this and the lowest possible value of (p - 3)² + 12 is 12.

This will be written as
(p - 3)² ≥ 0 for all real values of p
(p - 3)² + 12 ≥ 12 for all real values of p
Eric Nicholas K
Eric Nicholas K
2 years ago
We take advantage of the fact that a squared expression such as (p - 3)² will never have a negative value at all. It can only be zero or positive.

If (p - 3)² cannot go below 0, then of course (p - 3)² + 12 cannot go below 12.

How about the lowest possible values of things like (p - 3)² + 25 or (p - 3)² - 4? Of course they are 25 and -4 respectively, found by replacing p by 3 in each scenario.

For this to work, we must also ensure there are no other terms in p as well.