1
$\begingroup$

So I've gotten pretty far on this problem but I'm unable to complete it. I don't want to resort to the quadratic formula because this section of my review sheet specifically tells me not to use it. The problem is as follows:

$$3e^{2t} – e^t = 70$$

I substitute $y$ for $e^t$, and divide by 3, rewriting the equation is:

$$y^2 - \frac{y}{3} = \frac{70}{3}$$

But now I'm stuck. Factoring by grouping doesn't work and it doesn't seem like you can complete the square, could somebody help?

$\endgroup$
  • $\begingroup$ Why can't you complete the square? $\endgroup$ – Eric Towers Jun 4 '14 at 21:32
  • $\begingroup$ Completing the square actually works fine. You get a nice square on the right. $\endgroup$ – Daniel Fischer Jun 4 '14 at 21:33
  • $\begingroup$ You can always complete the square! :) $\endgroup$ – Just_a_fool Jun 4 '14 at 21:35
2
$\begingroup$

Here's an alternative to completing the square. Multiply both sides by 3 and collect all the terms on the left:

$$3y^2-y-70=0$$

Now look for two numbers whose product is $(3)(-70) = -210$ and whose sum is $-1$ (the latter from the coefficient of the $y$ term).

Those numbers must be nearly equal if their sum is $-1$, so they must be close to $\sqrt{210}$ in absolute value, which is around $14$ or $15$. And lo! the numbers $14$ and $-15$ fit the bill. Next:

$$3y^2 + 14y -15y -70 = 0$$ $$y(3y+14) - 5(3y+14) = 0$$ $$(y-5)(3y+14)=0$$

$\endgroup$
1
$\begingroup$

$$y^2 - \frac{y}{3} + \left( \frac{1}{2 \cdot 3} \right)^2 = \left(y - \frac{1}{6} \right)^2$$ $$ \frac{70}{3} + \left( \frac{1}{2 \cdot 3} \right)^2 = \left( \frac{29}{6} \right)^2 $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.