Quadratic Part1
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Notes
This is the one of the killer subtopics in the first half of the syllabus, second only to inequality. Just looking at it though, it doesn't seem like it, especially since we almost learn nothing new compared to add maths SPM. Two reasons :
- Anything that have already been covered in SPM, will have much more difficult questions in actual STPM, since it is pointless to issue SPM-level questions
- MOST students never actually did fully understand this topic in SPM. No, not even those with A1. Only a minority really understood the complexity of it. The rest just memorize formulas/workings.
Learning Objectives (Syllabus)
- use the process of completing the square for a quadratic polynomial
- derive the quadratic formula
- solve linear, quadratic, and cubic equations and equations that can be transformed into quadratic or cubic equations
- use the discriminant of a quadratic equation to determine the properties of its roots
- prove and use the relationship between the roots and coefficients of a quadratic equation
Prior Knowledge
- basic algebraic skills
- surds, complex numbers
- polynomials, factors, roots
Completing the Square
- This is one of the main skills that needs to be learnt well here. We will also need to master the various ways it can be used.
- I prefer to use a direct method of completing rather than using any formulas. It might seem difficult at first but you will get a hang of it.
Preliminary
Complete the following
Guideline
Similarly
Thus
Important
- Rearrange and factorize (if needed) so that
- Final form is
- CHECK!
Examples
Complete the square
Exercise 1
Complete the square for the following
a)
b)
c)
d)
e)
f)
To Find Max/Min
Once we have completed the square, we can obtain the max/min value of the polynomial just by looking at the final form. This, like the direct method of completing the square, requires logical thinking rather than memorizing of steps.
Value of x2
For all real values of
, we know that
That is,
Similarly
Meanwhile
That is,
Note
Since
for all real values of
,
Max/Min value
If
, it means
To Find Max/Min value
Note : MUST complete the square first (if not already so)
Max/Min Point
If maximum/minimum value of
when
maximum/minimum point is
Exercise 2
For every question in Exercise 1 (let the each polynomial to be equals
)
- a)State the maximum/minimum value of
and the value of
when it occurs
- b) the maximum/minimum point of its graph
a)
b)
c)
d)
e)
f)
Sign of Quadratic Polynomial
Preliminary
- If
,
- If
,
- If
- If
Thus
- If
has a positive minimum value
- or we say
- If
has a
- or we say
- If it is neither, but we are asked to prove always positive/negative, it means
To Show Sign of Quadratic Polynomial
From above, we can see that we would need to
Important Note:
Example
- Show that
for all
- And NOT
- Show that
is always negative.
To find Max/Min vs To Prove Sign
Suppose we have
, since
- If we need to find max/min,
- If we need to prove it is always positive,
Maximum/Minimum Value & Sign
- If given
- If given
Example
- Find range of
if
for all
Exercise 3
1) Prove that
for all
.
2) Prove that
for all
.
3) Prove that
for all
.
4) Prove that
is always positive for all
.
5) Find range of
if
for all
.
6) Given
for all
. Find range of
.
7) Show that if
for all
,
.






































on the left, we bring it over to the right and it becomes 









(need to multiply back if factored at the beginning)
(Write out the form first, including the
(so that we won't forget), and then only we fill in the blanks


gives us back


![={\color{Red}2\left[\left(x\qquad\right)^{2}\qquad\qquad-\frac{7}{2}\right]}](/images/math/8/f/5/8f5bcf63521ece16f11d8fad7c232c61.png)
![=2\left[\left(x{\color{Red}-\frac{3}{4}}\right)^{2}\qquad\qquad-\frac{7}{2}\right]](/images/math/8/b/e/8bed35fe20e0fca0f5875eb609b1679a.png)
![=2\left[\left(x-\frac{3}{4}\right)^{2}-{\color{Red}\frac{9}{16}}-\frac{7}{2}\right]](/images/math/a/5/2/a525ddd9f1d32ea23692e10b623e2500.png)

. Use calculator to check it will give back 

![=-\left[\left(x{\color{Red}-3}\right)^{2}\qquad+5\right]](/images/math/3/7/8/37838a9a28bd02b43114c9d33a93d9ff.png)
![=-\left[\left(x-3\right)^{2}{\color{Red}-9}+5\right]](/images/math/d/4/7/d4733bcc59c05115ecd98d3df17e60e5.png)


![\begin{align}
1-2x-3x^{2} & = -3x^{2}-2x+1\\
& = -3\left(x^{2}+\frac{2}{3}x-\frac{1}{3}\right)\\
& = -3\left[\left(x+\frac{1}{3}\right)^{2}-\frac{1}{9}-\frac{1}{3}\right]\\
& = -3\left(x+\frac{1}{3}\right)^{2}-\frac{4}{3}\\
\end{align}](/images/math/9/d/4/9d4514ef9ba2d2c6c57966e492508fc2.png)



![\begin{align}
2x^{2}-3x+11 & = 2\left(x^{2}-\frac{3}{2}x+\frac{11}{2}\right) \\
& = 2\left[\left(x-\frac{3}{4}\right)^{2}-\frac{9}{16}+\frac{11}{2}\right] \\
& = 2\left(x-\frac{3}{4}\right)^{2}+\frac{79}{8} \\
\end{align}](/images/math/8/b/2/8b226eb562bc88e276eac24b39fdc27e.png)

![\begin{align}
-3x^{2}+6x-3 & = -3\left(x^{2}-2x+1\right) \\
& = -3\left[\left(x-1\right)^{2}-1+1\right] \\
& = -3\left(x-1\right)^{2} \\
\end{align}](/images/math/0/c/8/0c856aa7acc9e97ee95d96dd0226a2ad.png)

![\begin{align}
1+6x-4x^{2} & = -4x^{2}+6x+1\\
& = -4\left(x^{2}-\frac{3}{2}x-\frac{1}{4}\right) \\
& = -4\left[\left(x-\frac{3}{4}\right)^{2}-\frac{9}{16}-\frac{1}{4}\right] \\
& = -4\left(x-\frac{3}{4}\right)^{2}+\frac{13}{4} \\
\end{align}](/images/math/0/6/e/06ea11d45fba49176297af1de9983706.png)

![\begin{align}
2x+7-3x^{2} & = -3x^{2}+2x+7 \\
& = -3\left(x^{2}-\frac{2}{3}x-\frac{7}{3}\right) \\
& = -3\left[\left(x-\frac{1}{3}\right)^{2}-\frac{1}{9}-\frac{7}{3}\right] \\
& = -3\left(x-\frac{1}{3}\right)^{2}+\frac{22}{3} \\
\end{align}](/images/math/e/b/d/ebd73ef967375bb0cc238d01ee9bf57c.png)








? Or
? It is
, so it still will be
is negative (when 







for all real values of
with 



















and then we will work from there.



(we can probably skip this step and straight to the max/min value step)
take this minimum value of 

(Can probably skip a step here)








?








take a minimum value of 











































, we need to prove that it is
, so instead of this line, we will write

![\begin{align}
2x-x^{2}-3 & = -x^{2}+2x-3 \\
& = -\left(x^{2}-2x+3\right) \\
& =-\left[\left(x-1\right)^{2}+2\right]\\
& =-\left(x-1\right)^{2}-2 \\
\end{align}](/images/math/0/2/5/0252ee611cd1bdb4cbd930f2ee482582.png)





(it must have a negative maximum value)

, so this value must be 








