Question

if the point(x, sqrt 3/2) is on the unit circle, what is x

Answer 1

The point is \[\Large \left(x, \frac{\sqrt{3}}{2}\right)\] right?

Answer 2

yes

Answer 3

Notice how that point is the form (x,y) where \[\Large y = \frac{\sqrt{3}}{2}\]
Do you agree with this statement?

Answer 4

yes

Answer 5

Now plug this y value into the unit circle equation and solve for x
\[\Large x^2+y^2 = 1\]
\[\Large x^2+\left( \frac{\sqrt{3}}{2}\right)^2 = 1\]
\[\Large x^2+ \frac{\left(\sqrt{3}\right)^2}{2^2} = 1\]
\[\Large x^2+ \frac{3}{4} = 1\]
Do you see how to solve for x from here?

Answer 6

no

Answer 7

how would you move the 3/4 to the other side? what must you do to both sides?

Answer 8

hint: you need to undo the addition

Answer 9

subtract 3/4 from each side

Answer 10

yes, that gives you

\[\Large x^2+ \frac{3}{4} = 1\]
\[\Large x^2+ \frac{3}{4}\color{red}{- \frac{3}{4}} = 1\color{red}{- \frac{3}{4}}\]
\[\Large x^2 = 1- \frac{3}{4}\]
\[\Large x^2 = \frac{4}{4}- \frac{3}{4}\]
\[\Large x^2 = \frac{4-3}{4}\]
\[\Large x^2 = \frac{1}{4}\]

Answer 11

Now if
\[\Large x^2 = \frac{1}{4}\]
then what must x be? How would you isolate x?

Answer 12

1/2

Answer 13

x = 1/2 is one solution

what is the other?

Answer 14

i don't know

Answer 15

hint: take a number like 10 and square it to get 100. You can also square -10 to get 100 as well since (-10)^2 = (-10)*(-10) = 100

so that's why the solutions to x^2 = 100 are x = -10 or x = 10