positive root of the equation $x^2+x-3-\sqrt{3}=0$

Question

$x^2+x-3-\sqrt{3}=0$ using the quadratic formula we get $$x=\frac{-1+\sqrt{13+4\sqrt{3}}}{2}$$ for the positive root

but the actual answer is simply $x=\sqrt3$

I am unable to perform the simplification any help would we helpful

Answer 1

$13+4\sqrt{3}=(2\sqrt{3})^2+2(2\sqrt{3})(1)+1=(2\sqrt{3}+1)^2$

So it is just $\dfrac{-1+2\sqrt{3}+1}{2}$.

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We may also start from the original equation.

\begin{align*} x^2+x-3-\sqrt{3}&=0\\ x^2-3+x-\sqrt{3}&=0\\ (x-\sqrt{3})(x+\sqrt{3})+x-\sqrt{3}&=0\\ (x-\sqrt{3})(x+\sqrt{3}+1)&=0 \end{align*}

Answer 2

Note that $(1+ 2 \sqrt3)^2 = 1 + 4 \sqrt 3 + 4 \cdot \sqrt 3^2 = 13 + 4 \sqrt 3$. The expression is now easy to simplify.

Answer 3

Let $a=\sqrt{3}$. \begin{align*} \text{Then}\;\;&x^2+x-3-\sqrt{3}=0\\[4pt] \iff\;&x^2+x=3+\sqrt{3}\\[4pt] \iff\;&x^2+x=a^2+a\\[4pt] \end{align*} So by inspection, we get the solution $x=a$.

By Vieta's formula, the sum of the roots is $-1$, so the other root is $-1-a$, which is negative.

Thus, $x=\sqrt{3}$ is the only positive root.

Alternatively, \begin{align*} &x^2+x=a^2+a\\[4pt] \iff\;&(x^2-a^2)+(x-a)=0\\[4pt] \iff\;&(x-a)(x+a+1)=0\\[4pt] \iff\;&x=a\;\,\text{or}\;\,x=-1-a\\[4pt] \iff\;&x=\sqrt{3}\;\,\text{or}\;\,x=-1-\sqrt{3}\\[4pt] \end{align*} so as before, $x=\sqrt{3}$ is the only positive root.

Answer 4

Rather than pulling $\sqrt{13+4\sqrt3}=1+2\sqrt3$ magically out of a hat, write $\sqrt{13+4\sqrt3}=a+b\sqrt3,$

as suggested by Parcly Taxel in a comment. Squaring both sides, $13+4\sqrt3=(a^2+3b^2)+2ab\sqrt3.$

$13=a^2+3b^2$ and $4=2ab$ means $b=2/a$ and $13=a^2+\dfrac{12}{a^2},$ i.e., $a^4-13a^2+12=0$, i.e., $(a^2-1)(a^2-12)=0$, i.e., $a=\pm1$ or $a=\pm2\sqrt3$ and respectively $b=\pm2$ or $\pm\dfrac 1{\sqrt3}. $ Note that $\sqrt{13+4\sqrt3}>0.$ Can you take it from here?