question in comments with correct math format

Question

anonymous · Answer

$$3\sqrt{2x-3}+2\sqrt{7-x}=11$$

unklerhaukus · Answer

square both sides

anonymous · Answer

so it would be: 3(2x-3)^2+2(7-x)^2=11^2

anonymous · Answer

whats after that? I am trying to check answers

unklerhaukus · Answer

actually this problem isn't as straight forward as i first thought

anonymous · Answer

oh i was wondering why I was super wrong.

anonymous · Answer

It is actually easier to bring one of the terms with the radical to the RHS to make expansion of squares easier.

anonymous · Answer

And doing so makes it straight forward :)

anonymous · Answer

would you mind showing me what you mean?

anonymous · Answer

$$3\sqrt{2x-3}=11-2\sqrt{7-x}$$

anonymous · Answer

$$(3\sqrt{2x-3})^2=(11-2\sqrt{7-x})^2$$
You have to isolate one of the radicals in an equation with multiple radicals

anonymous · Answer

Then you expand (x-y)^2=x^2-2xy+y^2

anonymous · Answer

And also, it is of important note to remember that square root and square are opposite operations and so they cancel

anonymous · Answer

so where did you get the x and y?

anonymous · Answer

so $$9(2x-3)=121-44\sqrt{7-x}+4(7-x)$$

anonymous · Answer

That is just to show you/remind you how to expand a polynomial.

anonymous · Answer

oh gotcha!

anonymous · Answer

:) Do you know where to go from there? and you would have to isolate your radical again.

theviper · Answer

$$(3\sqrt{2x-3}+2\sqrt{7-x})^2=11^2$$

theviper · Answer

Do that:)
Remember $$(a+b)^2=a^2+b^2+2ab$$

anonymous · Answer

yes thanks guys!

theviper · Answer

yw:)