Question

Solve for Q

Answer 1

\[ \sqrt{3q } + 2 = \sqrt{5 } \]

Answer 2

Begin by isolating the sqrt 3q:
\[\sqrt{3q}+2=\sqrt 5 \iff \sqrt{3q}=\sqrt5 -2 \]
Squaring both sides will help you get rid of the square root:
\[(\sqrt{3q})^2 = (\sqrt5-2)^2 \iff 3q=(\sqrt 5 -2)^2 \]

Answer 3

On the left hand-side you can see we plainly have a linear expression with the variable in question while in the right-hand side we have a squared binomial, which you can expand by using the formula: \((a+b)^2 = a^2 +2ab+b^2 \) where \(a,b \in \mathbb{R}\)

Answer 4

Oh okay so it shoudl look like this?
Q = \( {√5-2√3 } \)
*3 would be the denominator

Answer 5

Not quite, expanding the binomial, and it's a common mistake, is not the independent terms squared, this being: \[(\sqrt 5 -2)^2 = (\sqrt5)^2 +(2)(-2)(\sqrt5)+(-2)^2\]
Notice how I carried the "-2" since I can rewrite \((\sqrt5 -2)^2 \) as \(\left( \sqrt5 +(-2) \right)^2\).
Rather, you follow up the form, and you got the "3 in the denominator" correctly, since isolating "q" would imply we want to leave it alone on either side.

Answer 6

{√15-2√3 denominator is 3