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This is my attempt

$$ \begin{align} x^2 - mx &= 1 \\ x^2 - mx - 1 &= 0 \\ \left(x^2 - mx + \frac{m^2}{4} - \frac{m^2}{4}\right) - 1 &= 0 \\ \left(x^2 - mx + \frac{m^2}{4}\right) - \frac{m^2}{4} - 1 &= 0 \\ \left(x^2 - mx + \frac{m^2}{4}\right) - \frac{m^2}{4} - \frac{4}{4} &= 0 \\ \left(x^2 - mx + \frac{m^2}{4}\right) - \frac{m^2 - 4}{4} &= 0 \\ \left(x^2 - mx + \frac{m^2}{4}\right) &= \frac{m^2 - 4}{4} \\ \left(x - \frac{m}{2}\right)^2 &= \frac{m^2 - 4}{4} \\ \sqrt{\left(x - \frac{m}{2}\right)^2} &= \sqrt{\frac{m^2 - 4}{4}} \\ x - \frac{m}{2} &= \pm \frac{\sqrt{m^2 - 4}}{\sqrt{4}} \\ x &= \frac{m}{2} \pm \frac{\sqrt{m^2 - 4}}{2} \\[20pt] x_1 &= \frac{m}{2} - \frac{\sqrt{m^2 - 4}}{2} \\ x_1 &= \frac{m - \sqrt{m^2 - 4}}{2} \\[16pt] x_2 &= \frac{m}{2} + \frac{\sqrt{m^2 - 4}}{2} \\ x_2 &= \frac{m + \sqrt{m^2 - 4}}{2} \\ \end{align} $$

However, the correct answer according to the text is:

$$ \begin{align} x_1 &= \frac{m}{2} - \frac{\sqrt{m^2 + 4}}{2} \\ x_2 &= \frac{m + \sqrt{m^2 + 4}}{2} \\ \end{align} $$

Why $\sqrt{m^2 + 4}$ instead of $\sqrt{m^2 - 4}$ ???

pragMATHiC
  • 143

2 Answers2

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When you combined $$-\frac{m^2}{4} - \frac{4}{4}$$ into one fraction, you wrote it as $$-\frac{m^2-4}{4}.$$ You should have gotten $$-\frac{m^2+4}{4}$$ to make both terms appropriately negative.

Cameron Williams
  • 29,432
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it is $$x^2-2\frac{m}{2}x+\frac{m^2}{4}-\frac{m^2}{4}-1=0$$ and from here we get $$\left(x-\frac{m}{2}\right)^2=\frac{m^2}{4}+1$$ Can you finish?

Dr. Sonnhard Graubner
  • 95,283
  • $$ \begin{align} \left(x^2 - mx + \frac{m^2}{4}\right) &= \frac{m^2 + 4}{4} \ \left(x - \frac{m}{2}\right)^2 &= \frac{m^2 + 4}{4} \ \sqrt{\left(x - \frac{m}{2}\right)^2} &= \sqrt{\frac{m^2 + 4}{4}} \ x - \frac{m}{2} &= \pm \frac{\sqrt{m^2 + 4}}{\sqrt{4}} \ x &= \frac{m}{2} \pm \frac{\sqrt{m^2 + 4}}{2} \ x_1 &= \frac{m}{2} - \frac{\sqrt{m^2 + 4}}{2} \ x_1 &= \frac{m - \sqrt{m^2 + 4}}{2} \ x_2 &= \frac{m}{2} + \frac{\sqrt{m^2 + 4}}{2} \ x_2 &= \frac{m + \sqrt{m^2 + 4}}{2} \ \end{align} $$ – pragMATHiC May 10 '17 at 17:31