# Math Help - Simplification

1. ## Simplification

Have a go at this:

sqrt{7 - 4sqrt{3}} = 2 - sqrt{3}

How do I get from the left to the right hand side of the equation? When I got the answer to tan(15) and I typed it in, the RHS is what the calculator simplified it to. But HOW?!

2. Keep squaring on both sides until the roots are gone

3. Don't see how that's very relevant... I'm not solving for a variable.

4. Originally Posted by DivideBy0
Have a go at this:

sqrt{7 - 4sqrt{3}} = 2 - sqrt{3}

How do I get from the left to the right hand side of the equation? When I got the answer to tan(15) and I typed it in, the RHS is what the calculator simplified it to. But HOW?!
Square both sides:

the left hand side is:

7 - 4 sqrt(3),

and the rigth hand side is:

(2 - sqrt(3))(2-sqrt(3)) = 4 -4 sqrt(3) + 3 = 7 - 4 sqrt(3).

Now we need the tan 1/2 angle formula:

tan(A/2) = sqrt((1-cos(A))/(1+cos(A))

Now put A=30 degrees, then cos(30)=sqrt(3)/2, SO:

tan(15) = sqrt((1-sqrt(3)/2)/(1+sqrt(3)/2)) = sqrt((2-sqrt(3)/(2+sqrt(3))

but:

(2-sqrt(3)/(2+sqrt(3) = (2-sqrt(3))^2/[(2+sqrt(3))(2-sqrt(3))]

.............................= (4 -4 sqrt(3) +3)/(4-3) = 7-sqrt(3)

So:

tan(15) = sqrt(7 -sqrt(3)) = 2 - sqrt(3)

RonL

5. If you just had sqrt{7-4sqrt{3}}, with no other information, or knowledge that it was a half-angle, would it be possible to simplify?

6. Hello, DivideBy0!

Suppose we don't know the RHS . . .

Find: .sqrt[7 - 4·sqrt{3}]
Here is a primitive method for finding the square root.

. . . . . . . . . _ - . . ._______
Let: .a + b·√3 .= .√7 - 4·√3 . where a and b are integers
. . . . . . . . . . . . . . . . . . . ._ . . . . . ._______
Square both sides: .(a + b√3)² .= .(√7 - 4·√3)²
. . . . . . . . . . . . . . . . . ._ . . . . . . - . - . . ._
and we have: .a² + 2ab√3 + 3b² .= .7 - 4·√3
. . . . . . . . . . . . . . . . . . . _ . . . . . . . ._
Then: .(a² + 3b²) + (2ab)√3 .= .7 - 4·√3

Equate coefficients: . a² + 3b² .= .7 .[1]
. . . . . . . . . . . . . . . . . 2ab . . = -4 .[2]

And solve the system of equations . . .

From [2], we have: .b = -2/a .[3]

Substitute into [1]: .a² + 3(-2/a)² .= .7

. . which simplifies to: .a^4 - 7a² + 12 .= .0

. . which factors: .(a² - 3)(a² - 4) .= .0
. . . . . . . . . . . . . . . . . . . _
. . and has roots: .a .= .±√3, ±2

Since a is an integer, we will use: .a = 2

Substitute into [3]: .b = -2/2 = -1

. . . . . . . . . . . _______ . . . . . . . _
. . Therefore: .√7 - 4·√3 . = . 2 - √3

7. Hmm interesting, thanks for that

8. Originally Posted by DivideBy0
Hmm interesting, thanks for that
Note, Soroban's method does not always work.
This is mine 53th Post!!!

9. Just a quickie, for Soroban's method, why was +2 chosen over -2? How did you know it was the right one to use?

10. Hello, DivideBy0!

Why was +2 chosen over -2? How did you know it was the right one to use? .I didn't
It was laziness on my part . . . sorry.

Choosing a = +2, we get: b = -1, _
. . . and the square root is: . 2 - √3

Choosing a = -2, we get: b = 1, - -_
. . . and the square root is: .-2 + √3

As you suspected, there are two square roots for any quantity (except 0):
. . . . _______ . . . . . . . ._
. . . √7 - 4·√3 .= .±(2 - √3)