Friday, January 28, 2011

Math Recap 24-28/01/2011 Term 1

Done By: Tay Kun Yao (19) S2-05
This week, we've learnt about Quadratic Equations.

The General Formula of Quadratic
Equation is:

But first, we need to find out the value of "a, b and c".

To do so we have to rearrange the whole equation by making sure the equation is equals to 0. In other words, "3x - 1 = 6x^2 + 6x - 36" we've to deduct "3x - 1" and add "3x - 1" to the other side so that we would not change the equation. Thus, we would have an equation of "6x^2 + 3x - 35 = 0".

Therefore, we could figure out the values of "a, b and c" by using the formula:
where by a = 6, b = 3 and c = -35.

Now that we know the values of a, b and c. We can use the General Formula and solve for x by using the substituting method and get the answer of x which should look something like this:


To check if your answer(s) is/are correct, use the substitution method.

The value of a cannot be equivalent to 0.

If "square root of [(b)^2 - 4(a)(c)]" in the General Formula is:

A POSITIVE number, there are 2 Solutions.
A NEGATIVE number, there are No Solutions.
A PERFECT SQUARE, it is factorable
Lastly, If it is ZERO, there's 1 solution.

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