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A level Maths help!!!

how do you solve
8+x root12 = 8x/3

in the form a root b
Original post by x_tanzim
how do you solve
8+x root12 = 8x/3

in the form a root b


Collect terms, divide thorough, then multiply by '1' to remove the surf from the denominator.
Reply 2
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Original post by RogerOxon
Collect terms, divide thorough, then multiply by '1' to remove the surf from the denominator.

I still don't get it
Original post by x_tanzim
I still don't get it

Start by putting all you xes on one side, and all numbers on the other. Then simplify to get ax=b, where a and b may contain surds.
Original post by x_tanzim
I still don't get it

We can't post a solution as it's against the rules.

8+x root12 = 8x/3

8 = 8x/3 - x root 12

Now take out x on the right as a common factor and go from there.
Original post by x_tanzim
I still don't get it

8+x12=8x38+x\sqrt{12}=\frac{8x}{3}

∴8=x(83−12)\therefore 8=x(\frac{8}{3}-\sqrt{12})

Now continue - simplify the surd (hint: 12 = 3 * 4) and collect over a common denominator. Then multiply by the inverse of the coefficient of x. Post your working to that stage, and we'll go over the final step.
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Original post by RogerOxon
8+x12=8x38+x\sqrt{12}=\frac{8x}{3}

∴8=x(83−12)\therefore 8=x(\frac{8}{3}-\sqrt{12})

Now continue - simplify the surd (hint: 12 = 3 * 4) and collect over a common denominator. Then multiply by the inverse of the coefficient of x. Post your working to that stage, and we'll go over the final step.

I've simplified the surd to 2 root 3. Also you typed 8/3; it's 8/root 3
Original post by x_tanzim
I've simplified the surd to 2 root 3. Also you typed 8/3; it's 8/root 3

Not what you posted in your first post ....
Original post by x_tanzim
I've simplified the surd to 2 root 3. Also you typed 8/3; it's 8/root 3

OK, so now you have a coefficient of x: 83−23\frac{8}{\sqrt{3}} - 2\sqrt{3}

Can you place that all over 3\sqrt{3}?
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Original post by Muttley79
Not what you posted in your first post ....

okay that was a typo, it should be root 3
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Original post by RogerOxon
OK, so now you have a coefficient of x: 83−23\frac{8}{\sqrt{3}} - 2\sqrt{3}

Can you place that all over 3\sqrt{3}?

I honestly don't know. I've just started year 12 and I only just had my first proper maths lesson today. This is really confusing me.
Original post by x_tanzim
I honestly don't know. I've just started year 12 and I only just had my first proper maths lesson today. This is really confusing me.

Hint:

abc=abcb\frac{a}{b} c = \frac{a bc}{b}
(edited 4 years ago)
Original post by RogerOxon
Hint:

ab+c=a+bcc\frac{a}{b} + c = \frac{a + bc}{c}

I think you meant :smile:
ab+c=a+bcb\frac{a}{b} + c = \frac{a + bc}{b}
Reply 13
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Original post by muqeet137
I think you meant :smile:
ab+c=a+bcb\frac{a}{b} + c = \frac{a + bc}{b}

Yess I got the answerrr!!!
(edited 4 years ago)
Original post by muqeet137
I think you meant :smile:
ab+c=a+bcb\frac{a}{b} + c = \frac{a + bc}{b}

Yes - thanks.
Original post by x_tanzim
Yess I got the answerrr!!!

Good.

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