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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?

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    $\begingroup$ The discriminant might also be a perfect square $\endgroup$ – TokenToucan Feb 25 '16 at 7:11
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    $\begingroup$ 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
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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

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  • $\begingroup$ So what numbers would I substitute in to get the answers given? $\endgroup$ – bigfocalchord Feb 25 '16 at 7:14
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    $\begingroup$ @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

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