Question

A^2+X^2= sqrtx^2+4. Solve for A
WILL MEDAL AND FAN

Answer 1

you mean that there is a square root of left side,
or the square root (x^2+4) is really squared.

Answer 2

\[A^2+x^2=\sqrt{x^2+4}\]

Answer 3

A^2 + X^2 = X^2+4
A^2=4
A=2

that would be the adjacent side

(Assuming that this is the contiuation of the previous problem)

Answer 4

yes sir.

Answer 5

And so the adjacent side is 2, and hypotenuse is \(\sqrt{x^2+4}\), then sec would be equivalent to?

Answer 6

how does A = 2?

Answer 7

sec is equal to cos

Answer 8

No, sec is the reciprocal of cos

Answer 9

oops i was reading something else

Answer 10

and A=2, is what we had from:

A^2 + X^2 = X^2+4

Answer 11

secant = \(\rm Hypotenuse/Adjacent=\sqrt{x^2+4}/2\)

Answer 12

we are solving for "a" though. I thought we are suppose to single it out.

Answer 13

and same thing you would get from:

\(\large\color{black}{ \displaystyle \frac{1}{\sqrt{\dfrac{x^2+4}{x^2+4}-\dfrac{x^2}{x^2+4}}}=}\)


\(\large\color{black}{ \displaystyle \frac{1}{\sqrt{\dfrac{4}{x^2+4}}}=}\)


\(\large\color{black}{ \displaystyle \frac{1}{\dfrac{2}{\sqrt{x^2+4}}}=}\)


\(\large\color{black}{ \displaystyle \frac{\sqrt{x^2+4}}{2}.}\)

Answer 14

Can you do it where you would single out the "A"

Answer 15

I have solved for A, haven't I showed how?

Answer 16

you had that:

\(\large\color{black}{ \displaystyle {\rm A}^2+x^2=\left(\sqrt{x^2+4}\right)^2 }\)

Answer 17

i know you solved for A. but i dont understand how "A" becomes 2

Answer 18

\(\large\color{black}{ \displaystyle {\rm A}^2+x^2=\left(\sqrt{x^2+4}\right)^2 }\)

\(\large\color{black}{ \displaystyle {\rm A}^2+x^2=|x^2+4| }\)

assuming that x is real,

\(\large\color{black}{ \displaystyle {\rm A}^2+x^2=x^2+4 }\)

\(\large\color{black}{ \displaystyle {\rm A}^2=4\quad \Longleftarrow \quad {\rm A}=2 }\)

Answer 19

omg im so sorry i over complicated this lol. i got it now.

Answer 20

it's fine, \(....\)
(And same thing you will get for the secant, if you do what I initally offered.)

Answer 21

ok thanks a ton