Question: Find the three positive values of p for which the equation $$px^2-4x+1=0$$

will have rational roots.

My attempt (Algebraically):

Usually if it has to have rational roots then the discriminant must equal zero so

$$b^2 - 4ac = 0$$

$$ (-4)^4-4(p)(1)=0$$

$$ 16-4p = 0$$

$$ p = 4 $$

But the answers given are $3 , \frac{7}{4} , \frac{15}{4}$ how do they get that algebraically?

  • 1
    $\begingroup$ The discriminant might also be a perfect square $\endgroup$ – TokenToucan Feb 25 '16 at 7:11
  • 2
    $\begingroup$ The three positive values is a bit strange since there are infinitely many. $\endgroup$ – André Nicolas Feb 25 '16 at 7:18
  • $\begingroup$ I guess $p$ is supposed to be a positive integer. Then it makes perfect sense. $\endgroup$ – MooS Feb 25 '16 at 7:19
  • $\begingroup$ I agree with André Nicolas, the question is poorly phrased as there are infinitely made answers. $\endgroup$ – Ian Miller Feb 25 '16 at 7:19

To have rational roots , the discriminant must be perfect square

$$\implies16-4p=a^2\iff p=\dfrac{16-a^2}4$$ where $a$ is rational

If $p=0\iff a^2=16$ the equation won't have both roots finite

  • $\begingroup$ So what numbers would I substitute in to get the answers given? $\endgroup$ – bigfocalchord Feb 25 '16 at 7:14
  • 1
    $\begingroup$ @dydxx, You will have infinitely many solutions. You can substitute any rational number. They have substituted with $a=1,2,3$ $\endgroup$ – lab bhattacharjee Feb 25 '16 at 7:16
  • $\begingroup$ Thank you! I learnt something new today :) $\endgroup$ – bigfocalchord Feb 25 '16 at 7:17

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.