0
$\begingroup$

$$y^2-\frac25y=2$$

This is where I get:

$$\frac{\frac25 \pm\sqrt{(2/5)^2 - 4 \cdot 1 \cdot (-2)}}2$$ Within the square root part I get $8 +\frac4{25}$, I'm supposed to divide that by $2$. Supposedly I'm getting $\frac8{25}$ http://www.mathway.com/problem/NDcwMzE3OTIw

$$y=\frac{\dfrac25\pm\dfrac{2\sqrt{51}}{5}}{2\cdot 1}$$

Why did they get $\sqrt{51}$? I don't see how you can evaluate $(2/5)^2-4\cdot1\cdot(-2)$ and get that.

$\endgroup$
  • $\begingroup$ I am getting $y_{0,1} = \frac{\frac{2}{5}\pm \frac{13}{5}}{2}$. That is $y_0=\frac{3}{2}, y_1 = -1$. $\endgroup$ – quapka Dec 9 '14 at 21:11
  • 1
    $\begingroup$ @quapka That's wrong, the solutions are $$\frac15\left(1\pm\sqrt{51}\right)$$ $\endgroup$ – Alice Ryhl Dec 9 '14 at 21:15
  • $\begingroup$ @KristofferRyhl My bad, thanks. Calculated too fast and wanted the discriminant to be nice square, that's why I ended up with $\frac{4}{25} + 8 = \frac{169}{25}$, so off. :-) $\endgroup$ – quapka Dec 9 '14 at 21:48
1
$\begingroup$

Notice that it is not just $\sqrt{51}$, it is $\frac25\sqrt{51}$. You are right in getting $8+\frac{4}{25}=\frac{204}{25}$, but the given solutition takes this farther, factoring to get $\frac{204}{25}=51\cdot\frac{4}{25}=51\cdot\left(\frac{2}{5}\right)^2$. This means that $\sqrt{\frac{204}{25}}=\sqrt{51\cdot\left(\frac{2}{5}\right)^2}=\frac{2}{5}\sqrt{51}$.

$\endgroup$
0
$\begingroup$

Clear denominator: $5y^2 - 2y - 10 = 0 \Rightarrow y = \dfrac{-(-2) \pm \sqrt{(-2)^2 - 4\cdot 5\cdot (-10)}}{2\cdot 5} = \dfrac{2 \pm \sqrt{204}}{10} = \dfrac{2 \pm 2\sqrt{51}}{10} = \dfrac{1 \pm \sqrt{51}}{5}$.

$\endgroup$

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.