Math Help - squaring brackets

1. squaring brackets

i was wondering what the method is to figuring this out:

$(\frac {36-4}{8})^2$ am i right in thinking the answer is 16?

2. Originally Posted by mark
i was wondering what the method is to figuring this out:

$(\frac {36-4}{8})^2$ am i right in thinking the answer is 16?
Hi mark

$\left(\frac{36-4}{8}\right)^2=\left(\frac{32}{8}\right)^2=(4)^2=1 6$

3. ah thanks masters, so i've got that part of it right then. do you reckon you could look at the last question i posted on here please? if you could could you explain everything i need to do to complete it in detail becase there must be something i'm missing thats making it hard to understand. thanks

ps its the one called "equation"

4. Originally Posted by mark
ah thanks masters, so i've got that part of it right then. do you reckon you could look at the last question i posted on here please? if you could could you explain everything i need to do to complete it in detail becase there must be something i'm missing thats making it hard to understand. thanks

ps its the one called "equation"
Sure, mark,

(1) $5x + 3y = 4$
(2) $5x^2 - 3y^2 = 8$

Solve the first equation for x.

(1) $x=\frac{4-3y}{5}$

Substitute this in (2).

$5\left(\frac{4-3y}{5}\right)^2-3y^2=8$

Expand and simplify:

$5\left(\frac{16-24y+9y^2}{25}\right)-3y^2=8$

$\left(\frac{16-24y+9y^2}{5}\right)-3y^2=8$

Multiply all terms by 5.

$16-24y+9y^2-15y^2=40$

Combine terms and simplify.

$-6y^2-24y-24=0$

$y^2+4y+4=0$

Factor.

$(y+2)(y+2)=0$

$y=-2$

Now, substitute this back into (1) to find the corresponding x value.

$5x+3(-2)=4$

$5x-6=4$

$5x=10$

$x=2$

The intersection is (2, -2). The line is tangent to the hyperbola at this point.

5. ahhh thankyou very much, that was a really good explanation