1. ## All igcse questions

hey everybody
i couldnt figure out this question from a past paper about quadratic formula
it was pretty straight foward-
the total area of rectangle R and Q is 64cm^2
calculate the value of x correct to one decimal place
5x^2+30x+24
i got 0.95 (for subtraction) and 5.05

i am jus confused on what i am suppose to choose so that the answer is 64
plus forgot what it meant by correct to one decimal place...was it rounding up? ...n also 3 significant figures....actually i m mixed up in both of them -.-

thanks

2. Originally Posted by darkpinkish
hey everybody
i couldnt figure out this question from a past paper about quadratic formula
it was pretty straight foward-
the total area of rectangle R and Q is 64cm^2
calculate the value of x correct to one decimal place
5x^2+30x+24
i got 0.95 (for subtraction) and 5.05

i am jus confused on what i am suppose to choose so that the answer is 64
plus forgot what it meant by correct to one decimal place...was it rounding up? ...n also 3 significant figures....actually i m mixed up in both of them -.-

thanks
From what I can gather 64 is what the quadratic function is equal to.

$5x^2-30x+24=64$ which is equal to $(x-2)(x-4)=0$

I have no idea how you got the quadratic

3. same here...but the asnwer is 1.1,
from the first part of the question i've arrived to 5x^2+30x+24
and the second part said find x correct 1 decimal place and the total area equals 64cm^2
i got the answer as 0.95 and 5.05
i was wondering if 0.95 corrected to one decimal place is 1.1, since i m confused between corrct to 1 decimal place and correct to 1 significant figure

4. Originally Posted by darkpinkish
hey everybody
i couldnt figure out this question from a past paper about quadratic formula
it was pretty straight foward-
the total area of rectangle R and Q is 64cm^2
calculate the value of x correct to one decimal place
5x^2+30x+24
Surely the "past paper" you got this from must have some information about "Rectangle R and Q" and what they have to do with $5x^2+ 30x+ 24$!

As stated they don't appear to have anything to do with one another and there is no way to find "x" with just a polynomial and not an equation.

i got 0.95 (for subtraction) and 5.05

i am jus confused on what i am suppose to choose so that the answer is 64
plus forgot what it meant by correct to one decimal place...was it rounding up? ...n also 3 significant figures....actually i m mixed up in both of them -.-

thanks

5. ## Functions

Find ff^-1(8)

and i even want to know how is it different from f^-1(x) , the inverse

=)

6. Originally Posted by darkpinkish
Find ff^-1(8)

and i even want to know how is it different from f^-1(x) , the inverse

=)
$ff^{-1}(8) = 8$. This is because a function and it's inverse cancel each other out.

$ff^{-1}(x)$ means substitute the inverse of $f(x)$ into the original $f(x)$. However, because these two are "opposite" functions the answer is always x.

See spoiler for examples:

Spoiler:
Example 1:

$f(x) = \frac{a}{x} \, , \: a,x \neq 0$ then $f^{-1}(x) = \frac{a}{x}$.

Therefore $ff^{-1}(x) = \frac{a}{\frac{a}{x}} = x$. This follows because $f(x) = f^{-1}(x)$

Example 2:

$f(x) = x^2 \, , \: x \geq 0$

$y = x^2 \: \rightarrow \: x = \sqrt{y}$

$\therefore f^{-1}(x) = \sqrt{x} \: , \: x \geq 0$

$ff^{-1}(x) = (\sqrt{x})^2 = x$

If you have any more questions on functions and their inverses I suggest you start a new thread