How do you prove that $\sqrt{4 + 2 \sqrt{3}} = \sqrt{3} + 1$ $sqrt(4+2sqrt(3)) = sqrt(3)+1$ ?

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How do you prove that $\sqrt{4 + 2 \sqrt{3}} = \sqrt{3} + 1$ $sqrt(4+2sqrt(3)) = sqrt(3)+1$ ?

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Answer 1 · 2016-11-20T02:20:56

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 $1$ $1$ . Beginning with the quotient, we have

$\frac{\sqrt{4 + 2 \sqrt{3}}}{\sqrt{3} + 1}$ $sqrt(4+2sqrt(3))/(sqrt(3)+1)$

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

$\frac{\sqrt{4 + 2 \sqrt{3}}}{\sqrt{3} + 1} = \frac{\sqrt{4 + 2 \sqrt{3}} \times (\sqrt{3} - 1)}{(\sqrt{3} + 1) \times (\sqrt{3} - 1)}$ $sqrt(4+2sqrt(3))/(sqrt(3)+1) = (sqrt(4+2sqrt(3))xx(sqrt(3)-1))/((sqrt(3)+1)xx(sqrt(3)-1))$

$= \frac{\sqrt{4 + 2 \sqrt{3}} \times (\sqrt{3} - 1)}{{(\sqrt{3})}^{2} - 1^{2}}$ $=(sqrt(4+2sqrt(3))xx(sqrt(3)-1))/((sqrt(3))^2-1^2)$

$= \frac{\sqrt{4 + 2 \sqrt{3}} \times (\sqrt{3} - 1)}{3 - 1}$ $=(sqrt(4+2sqrt(3))xx(sqrt(3)-1))/(3-1)$

$= \frac{\sqrt{4 + 2 \sqrt{3}} \times (\sqrt{3} - 1)}{2}$ $=(sqrt(4+2sqrt(3))xx(sqrt(3)-1))/2$

As the quotient must be equal to $1$ $1$ if the given expressions are equal, we now need to show that the numerator is equal to $2$ $2$ .

$\sqrt{4 + 2 \sqrt{3}} \times (\sqrt{3} - 1) = \sqrt{4 + 2 \sqrt{3}} \times \sqrt{{(\sqrt{3} - 1)}^{2}}$ $sqrt(4+2sqrt(3))xx(sqrt(3)-1) = sqrt(4+2sqrt(3))xxsqrt((sqrt(3)-1)^2)$

(Note that the above step is justified because $\sqrt{3} - 1 > 0$ $sqrt(3)-1 > 0$ . If $x \geq 0$ $x>=0$ , then $x = \sqrt{x^{2}}$ $x = sqrt(x^2)$ . If $x < 0$ $x<0$ , then $x = - \sqrt{x^{2}}$ $x=-sqrt(x^2)$ )

$= \sqrt{(4 + 2 \sqrt{3}) {(\sqrt{3} - 1)}^{2}}$ $=sqrt((4+2sqrt(3))(sqrt(3)-1)^2)$

$= \sqrt{(4 + 2 \sqrt{3}) (3 - 2 \sqrt{3} + 1)}$ $=sqrt((4+2sqrt(3))(3-2sqrt(3)+1))$

$= \sqrt{(4 + 2 \sqrt{3}) (4 - 2 \sqrt{3})}$ $=sqrt((4+2sqrt(3))(4-2sqrt(3))$

$= \sqrt{4^{2} - {(2 \sqrt{3})}^{2}}$ $=sqrt(4^2-(2sqrt(3))^2)$

$= \sqrt{16 - 12}$ $=sqrt(16-12)$

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

$= \sqrt{4}$ $=sqrt(4)$

$= 2$ $=2$

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

$\frac{\sqrt{4 + 2 \sqrt{3}}}{\sqrt{3} + 1} = \frac{\sqrt{4 + 2 \sqrt{3}} \times (\sqrt{3} - 1)}{(\sqrt{3} + 1) \times (\sqrt{3} - 1)} = \frac{2}{2} = 1$ $sqrt(4+2sqrt(3))/(sqrt(3)+1) = (sqrt(4+2sqrt(3))xx(sqrt(3)-1))/((sqrt(3)+1)xx(sqrt(3)-1))=2/2=1$

$\Rightarrow \frac{\sqrt{4 + 2 \sqrt{3}}}{\sqrt{3} + 1} \times (\sqrt{3} + 1) = 1 \times (\sqrt{3} + 1)$ $=> sqrt(4+2sqrt(3))/(sqrt(3)+1)xx(sqrt(3)+1) = 1xx(sqrt(3)+1)$

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

Answer 2 · 2016-11-20T12:36:51

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

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

Answer 3 · 2016-11-20T12:59:28

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:

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

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