Grade 10 math, radicals (?)

Printable View

April 20th 2009, 02:13 PM
Tommy.

Grade 10 math, radicals (?)

Hello ive worked these out in my notebook but havbe no way to check if they are correct or not here they are

a) ( √2 - √8)² = 2
b) ( -1/3√3 - 1 )²
c) ( √3 + √12 ) ²

I know I got to do the perfect square trinomial
a²+ 2ab + b² I did that just would like the working out of it and answers in case I got them wrong

ps , im not skipping on my homewrok just got a test on thurdays and want to see if im ok since i go t no more math classes till then thats why I ask for the work out of it

THANKSS
April 20th 2009, 02:28 PM
pickslides

Tommy your answer for a) is correct.

For the other questions

$b) \frac{4}{3} + \frac{2}{3} \sqrt{3}$

$c) 27$

You are correct in saying you need to use a perfect square binomial expansion.
April 20th 2009, 02:33 PM
Tommy.

can I see how a b and c are done? if its not too much trouble for anyone (Lipssealed)

got a right but B and C wrong (Worried)
April 20th 2009, 02:44 PM
pickslides

question c)

$(\sqrt{3} + \sqrt{12})^2$

$= (\sqrt{3} + \sqrt{12}) (\sqrt{3} + \sqrt{12})$

now multiply the terms out using this general form.

$= (a + b) (c + d)$

$= ac + ad + bc + bd$

this will given you

$= \sqrt{3} \sqrt{3}+ \sqrt{3} \sqrt{12} + \sqrt{12} \sqrt{3} + \sqrt{12}\sqrt{12}$

Then simplifying gives

$= 3+ 2 \sqrt{3} \sqrt{12} +12$

$= 3+ 2 \sqrt{36} +12$

$= 3+ 2 .6 +12$

$= 3+ 12 +12$

$= 27$

Follow similar steps for question b)
April 20th 2009, 04:02 PM
Tommy.

Find value of X with radicals

a) √2 (x - √3) + 5 √6 = - √6 / 2

b) x+√5 (minus) X- √15 = 0 * both of the terms have a 2 as denominator
----- --------
2 . 2

c) (x + 2√10) (x - √90) = (cubed root)3√36

---

This is a new unit lets say and i want to see if I can get ahead a little, having the thing worked out in full would be awsome so I have some background as to what our first class after the test will be taught
April 20th 2009, 04:16 PM
Tommy.

Find value of X with radicals

Double post, wasnt a bump im in argentina and these things get messy
April 20th 2009, 05:18 PM
pickslides

Quote:

Originally Posted by Tommy.

a) √2 (x - √3) + 5 √6 = - √6 / 2

b) x+√5 (minus) X- √15 = 0 * both of the terms have a 2 as denominator
----- --------
2 . 2

c) (x + 2√10) (x - √90) = (cubed root)3√36

---

This is a new unit lets say and i want to see if I can get ahead a little, having the thing worked out in full would be awsome so I have some background as to what our first class after the test will be taught

so the new problems are:

a) $\sqrt{2} (x-\sqrt{3}) + 5\sqrt{6} = \frac{-\sqrt{6}}{2}$

b) $\frac{x+\sqrt{5}}{2} - \frac{x-\sqrt{5}}{2} = 0$

c) $(x+2\sqrt{10})(x -\sqrt{90}) = \sqrt[3]{36}$
April 21st 2009, 06:04 PM
Tommy.

A and C are ok , this is how B should be

b) $\frac{x+\sqrt{5}}{2} - \frac{x-\sqrt{15}}{2} = 0$
April 21st 2009, 07:11 PM
pickslides

Quote:

Originally Posted by Tommy.

A and C are ok , this is how B should be

b) $\frac{x+\sqrt{5}}{2} - \frac{x-\sqrt{15}}{2} = 0$

Are you sure this is correct?

$\frac{x+\sqrt{5}}{2} - \frac{x-\sqrt{15}}{2} = 0$

I would multiply both sides through by 2 giving

$(x+\sqrt{5})-(x-\sqrt{15}) = 0$

then,

$x+\sqrt{5}- x+\sqrt{15} = 0$

And now the x’s cancel and gives the following non-sensical statement as

$\sqrt{5} +\sqrt{15} \neq 0$