How do you prove that sqrt(4+2sqrt(3)) = sqrt(3)+14+23=3+1 ?

3 Answers
Nov 20, 2016

One way to show that the left hand side is equal to the right hand side is to show that their quotient is equal to 11. Beginning with the quotient, we have

sqrt(4+2sqrt(3))/(sqrt(3)+1)4+233+1

To help us evaluate this, let's first rationalize the denominator

sqrt(4+2sqrt(3))/(sqrt(3)+1) = (sqrt(4+2sqrt(3))xx(sqrt(3)-1))/((sqrt(3)+1)xx(sqrt(3)-1))4+233+1=4+23×(31)(3+1)×(31)

=(sqrt(4+2sqrt(3))xx(sqrt(3)-1))/((sqrt(3))^2-1^2)=4+23×(31)(3)212

=(sqrt(4+2sqrt(3))xx(sqrt(3)-1))/(3-1)=4+23×(31)31

=(sqrt(4+2sqrt(3))xx(sqrt(3)-1))/2=4+23×(31)2

As the quotient must be equal to 11 if the given expressions are equal, we now need to show that the numerator is equal to 22.

sqrt(4+2sqrt(3))xx(sqrt(3)-1) = sqrt(4+2sqrt(3))xxsqrt((sqrt(3)-1)^2)4+23×(31)=4+23×(31)2

(Note that the above step is justified because sqrt(3)-1 > 031>0. If x>=0x0, then x = sqrt(x^2)x=x2. If x<0x<0, then x=-sqrt(x^2)x=x2)

=sqrt((4+2sqrt(3))(sqrt(3)-1)^2)=(4+23)(31)2

=sqrt((4+2sqrt(3))(3-2sqrt(3)+1))=(4+23)(323+1)

=sqrt((4+2sqrt(3))(4-2sqrt(3))=(4+23)(423)

=sqrt(4^2-(2sqrt(3))^2)=42(23)2

=sqrt(16-12)=1612

(As when rationalizing the denominator, we make use of the identity (a+b)(a-b) = a^2-b^2(a+b)(ab)=a2b2)

=sqrt(4)=4

=2=2

Now that we have shown that the numerator has the desired property, we can solve the rest of the problem quite simply.

sqrt(4+2sqrt(3))/(sqrt(3)+1) = (sqrt(4+2sqrt(3))xx(sqrt(3)-1))/((sqrt(3)+1)xx(sqrt(3)-1))=2/2=14+233+1=4+23×(31)(3+1)×(31)=22=1

=> sqrt(4+2sqrt(3))/(sqrt(3)+1)xx(sqrt(3)+1) = 1xx(sqrt(3)+1)4+233+1×(3+1)=1×(3+1)

:. sqrt(4+2sqrt(3)) = sqrt(3)+1

Nov 20, 2016

See below.

Explanation:

This expression has the structure

sqrt(a+bsqrt(3))=csqrt(3)+d so squaring both sides

a+bsqrt(3)=3 c^2+ 2 sqrt[3] c d + d^2 pairing terms

{(a - 3 c^2 - d^2 = 0),(b - 2 c d=0):}

Solving for c,d we have

c = pmsqrt[a pm sqrt[a^2 - 3 b^2]]/sqrt[6]
d=pm(sqrt[3/2] b)/sqrt[a - sqrt[a^2 pm 3 b^2]]

If a=4, b=2 we have the possibilities

((c= -1/sqrt[3], d= -sqrt[3]),(c= 1/sqrt[3], d= sqrt[3]),(c= -1, d = -1),(c= 1, d = 1))

Nov 20, 2016

See description...

Explanation:

Note that:

(sqrt(3)+1)^2 = (sqrt(3))^2+2(sqrt(3))+1

color(white)((sqrt(3)+1)^2) = 3+2 sqrt(3)+1

color(white)((sqrt(3)+1)^2) = 4+2sqrt(3)

Since sqrt(3)+1 > 0, we can take the positive square root of both ends to get:

sqrt(3)+1 = sqrt(4+2sqrt(3))