<s>[INST] Help with a geometry problem The problem says:
A triangle has its lengths in an arithmetic progression,
with difference d. The area of the triangle is t. Find the dimensions.
the solution says:
the notation can be even better if we make it more symmetrical, by making the
side lengths  $b − d, b,$ and $b + d$ .
by Heron’s formula we know that $t^2 = s(s − b + d)(s − b)(s − b − d)$ , where $s =
((b − d) + b + (b + d))/2$ is the semi-perimeter;
and after simplification 
$$ t^2 = \frac{3b}{2}(\frac{3b}{2} - b + d )(\frac{3b}{2} - b ) (\frac{3b}{2} - b - d )  $$
$$ \implies t^2 = \frac{3b^2(b-2d)(b+2d)}{16} = \frac{3b^2(b^2-4d^2)}{16} $$
then we get $$ 3b^4 − 12d^2b^2 − 16t^2 = 0 $$
and using the quadratic formula :
$$ b^2 = \frac{12d^2 \pm \sqrt{144d^4 + 169t^2} }{6} = 2d^2 \pm \sqrt{4d^4 + \frac{16}{3} t^2} $$
and because b has to be positive , we get
$$ b = \sqrt{2d^2 + \sqrt{4d^4 + \frac{16}{3}t^2}} $$
Which is the part that i have a problem with , my question is :
why should we select only the positive sign solution of the quadratic formula ? is that because $\sqrt{ 4d^4 + \frac{16}{3}t^2} > 2d^2$ which means that the negative sign solution leads to the square root of a negative number which is not valid? why is the positive sign solution is the right solution ?
In other words :
If  $\sqrt{ 4d^4 + \frac{16}{3}t^2} > 2d^2$  , how is that ? how can we prove it ?
thank you
 [/INST] We have $4d^4=(2d^2)^2$, and therefore $4d^4+\frac{3}{16}t^2>(2d^2)^2$ and therefore 
$$\sqrt{4d^4+\frac{3}{16}t^2}>2d^2.$$
 It follows that $2d^2-\sqrt{4d^4+\frac{3}{16}t^2}<0$, so  $2d^2-\sqrt{4d^4+\frac{3}{16}t^2}$ does not have a (real) square root.
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