# Help me with my grade 11 maths [easy maths]

• Dec 27th 2009, 01:09 PM
Help me with my grade 11 maths [easy maths]
It is surely a piece of cake for you guys, but I just can't grasp how to do it. It's the first time in 11 years that I actually have to study math. Every year was easy, except this one. I'm the one to be blamed for not studying, but in 11 years, I haven't had such a bad teacher. Anyways...

I have a hard time resolving functions (I think that's what's called... "resolution des fonctions" in french.).
We did quadratic functions (ax² + 2bx + c) last year, and they're quite easy. To find the zeros of the function, all you need is the following formula ...
(-b ± √(b²‎ - 4ac)) / 2a
What about square root functions, rational functions, absolute value functions, simple functions? What formulas do they use? If they do not use formulas, how is it done? I would really appreciate it if you guys give me an example of how it's done if you have f(x) = 0 , f(x) > 5, f(x) < 5 (inequations)

Also, another thing that has been confusing me - the teacher often mixes two functions together. How do I solve that?
For example - square root = absolute value
3√(2x + 6) + 3 = 2|4x-2| + 2

I do realize how easy it is. I can't really do anything though - the teacher isn't really helpful, and being in the academic reform, the schoolbooks aren't ready yet, so there is nowhere I can study from. I am sorry that I'm asking for so much, but it's a piece of cake math for most of you, so any help is greatly, greatly appreciated.

• Dec 27th 2009, 02:54 PM
NOX Andrew
Solving algebraic equations involves reversing (undoing) the order of operations. In order of precedence, the order of operations are parentheses, exponentiation, multiplication and division, and addition and subtraction. The order of operations can be remembered by the mnemonic PEMDAS. The reverse of multiplication is division (and vice versa) and the reverse of addition is subtraction (and vice versa).

I'll use the following equation as an example.

$2x + 5 = 0$

To solve the above equation in terms of $x$, reverse the order of operations. First, subtract 5 on both sides of the equation to reverse the addition.

$2x + 5 - 5 = 0 - 5$

$2x = -5$

As you can see, the addition was undone by the subtraction. Now, divide by 2 on both sides of the equation to reverse the multiplication.

$\frac{2x}{2} = \frac{-5}{2}$

$x = \frac{-5}{2}$

Reversing exponentiation involves a similar process. Take the following equation for example.

$x^\frac{1}{2} = 3$

To reverse exponentiation, raise each side of the equation to the reciprocal power. The power is $\frac{1}{2}$ so the reciprocal is $\frac{2}{1} = 2$. Therefore, raise each side of the equation to 2nd power.

$(x^\frac{1}{2})^2 = 3^2$

Remember the Laws of Exponents, specifically $(x^a)^b = x^{ab}$. Therefore, $(x^\frac{1}{2})^2 = x^{\frac{1}{2} \times 2} = x^{\frac{1}{2} \times \frac{2}{1}} = x^\frac{2}{2} = x^1 = x$. As you can see, this undoes the exponentiation. Thus, we have:

$x = 3^2$

$x = 9$

Edit: Yes, nearly all if not all algebraic equations can be solved by undoing the order of operations.
• Dec 27th 2009, 02:56 PM
Thanks a lot. Is this true for ALL the functions however?
• Dec 27th 2009, 02:57 PM
Bacterius
Note that an absolute value can be removed by considering the two cases (expression is negative or positive), and sometimes you can totally get rid of it by squaring.

EDIT : resolving equations always implies reversing the operations done to obtain it.
• Dec 27th 2009, 03:14 PM
NOX Andrew
I followed the link you provided for simple functions and realized these require different methods.

I'll use floor() to denote the flooring function. If you aren't familiar with the flooring function, it rounds a real number down to the nearest integer. For example, floor(5.5) = 5 and floor(-3.7) = -4.

Here is an example equation:

$floor(x) = 12$

A lot of numbers are solutions to this equation. 12, 12.123, 12.333..., and 12.99 to name a few. In fact, all numbers between 12 and 13 (including 12 but excluding 13) are solutions to the equation. Such a solution can be mathematically expressed as $12 \le x < 13$. Different simple functions require different methods. They are more conceptual than analytical.
• Dec 27th 2009, 03:18 PM
Raoh
hi(Happy)
Remember the following properties,
$\forall a\in \mathbb{R}$ you have :
$\left | a \right |^2=a^2$.
$\sqrt{a^2}=\left | a \right |$
$\left | a \right |=a$ if $a\geq 0$.
$\left | a \right |=-a$ if $a< 0.$
(by the way "Résolution des équations = solving equations (Happy) ).
• Dec 27th 2009, 03:46 PM
Thanks a lot guys for the replies. All of them have been helpful.
I still am clueless how to do a system of equations, though.

For example, ...

y1 = 3√(2x + 6) + 3
y2 = 2|4x-2| + 2
y1 = y2 if x = ?

Thanks :D
• Dec 27th 2009, 03:48 PM
Bacterius
In a system of equations, you try to express one unknown with another with one of the equations, and then you can substitute into another equation to solve.

You can think of it as quantity of information : a system of two equations of two unknowns holds enough information to allow solving for both, provided you can use the information.

EDIT : ah, didn't see your system right then. That's not my definition of a system, though. The question suggests you to let $3 \sqrt{2x + 6} + 3 = 2 |4x-2| + 2$ and solve for $x$. Is that where you fail ?
• Dec 27th 2009, 04:02 PM
It's called a systeme d'equations in French, sorry if my translation was inadequate.

Yeah, that's where I fail. Basically, for what http://www.mathhelpforum.com/math-he...155c67a6-1.gif does http://www.mathhelpforum.com/math-he...b440b6d8-1.gif?
• Dec 27th 2009, 04:02 PM
Raoh
Here are the types of inequalities that you should know :
Type 1 :
$\sqrt{f(x)}< g(x)$
$\sqrt{f(x)}< g(x)\Leftrightarrow \left (f(x)\geq 0,g(x)\geq 0,f(x)< g^2(x) \right )$
Type 2 :
$\sqrt{f(x)}< \sqrt{g(x)}$
$\sqrt{f(x)}< \sqrt{g(x)}\Leftrightarrow \left (f(x)\geq 0,f(x)< g(x) \right )$
Type 3 :
$f(x)\leq \sqrt{g(x)}$
$f(x)\leq \sqrt{g(x)}\Leftrightarrow$ $\left (\left \{f(x)\leq 0,g(x)\geq 0 \right \} \text{or} \left \{ f(x)\geq 0,f^2(x)\leq g(x) \right \} \right )$
• Dec 27th 2009, 04:07 PM
Bacterius
No problem for your translation, I am french myself.
Actually, it is a system of equations, but it is already substituted and ready to solve.
Did you try graphing the function to see for which $x$ it works ?
• Dec 27th 2009, 04:09 PM
Raoh
Quote:

and don't forget you also have $2x+6\geq 0$.