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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

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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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Yes, you know that $b^2$ and $b$ are both positive real numbers, so you select the positive sign twice while taking square root.

If you want to know why the inequality is true, note that both sides of it are positive, so you can just square both sides to get an equivalent inequality which is easily seen to be correct.

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