Extending Completing the square to expressions of $ (x+y+z)^2$

Question

We all know expressions such as $x^2+14x+49 = (x+7)^2$ because it is easily recognizable as a perfect square. What about a expressions in the form of $x^2+2xy+2yz+2xz+y^2+z^2=(x+y+z)^2$

The question is how do I factor $0.09e^{-2t}+0.24e^{-t}+0.34+0.24e^t+0.09e^{2t}$ into the form of $(x+y+z)^2$?

Answer 1

$(0.3e^{-t})^2+0.24e^{-t}+0.4^2+2*0.3*0.3+0.24e^t+(0.3e^t)^2=$$(0.3e^{-t}+0.4+0.3^t)^2$

Answer 2

The answer will certainly be $(ae^t + b + ce^{-t})^2$ for some real $a, b, c$. Expand it, group terms by the power of $e$ and match them with the constants in your formula. You will get a system of equations you need to solve.

Answer 3

HINT : Pay attention to $0.09e^{\pm 2t}=(0.3e^{\pm t})^2$. Then, you'll have the following form as $$(0.3e^t+0.3e^{-t}+\alpha)^2$$

Answer 4

Setting $e^t=a,$ we have $\displaystyle\frac{9+24a+34a^2+24a^3+9a^4}{100a^4}$ which can be written as $\displaystyle\frac{(3a^2+ab+3)^2}{(100a^2)^2}$ where $b$ is independent of $a$

$\displaystyle\implies 9+24a+34a^2+24a^3+9a^4=9a^4+a^2b^2+9+6ba^3+6ba+18a^2$

$\displaystyle\implies 9+24a+34a^2+24a^3+9a^4=9a^4+6ba^3+a^2(18+b^2)+6ba+9$

Comparing the coefficients of $a$ or $a^3,\displaystyle24=6b\iff b=4$

Comparing the coefficients of $a^2,\displaystyle34=18+b^2\iff b^2=16\iff b=\pm4$

So, $b=?$