# Find the three positive values of p for which the equation $px^2-4x+1=0$ will have rational roots

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?

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

$$\implies16-4p=a^2\iff p=\dfrac{16-a^2}4$$ where $a$ is rational
If $p=0\iff a^2=16$ the equation won't have both roots finite
• @dydxx, You will have infinitely many solutions. You can substitute any rational number. They have substituted with $a=1,2,3$ – lab bhattacharjee Feb 25 '16 at 7:16