Answer:
- [tex]\mathrm{No\:Solution}[/tex]
Step-by-step explanation:
- given that: [tex](p^2 +1)x^2 +2pqx+q^2 = 0[/tex]
- also its being said that: it has no solution
- thus shall we use the formula: b²-4ac < 0
Extra's:
- when it has solution: b²-4ac > 0
- when has only one or equal use: b²-4ac = 0
- when it has no solution or real numbers: b²-4ac < 0
Now lets solve:
[tex](p^2 +1)x^2 +2pqx+q^2 = 0[/tex]
[tex](p^2 +1)x^2 +(2pq)x+q^2 = 0[/tex]
using b²-4ac < 0:
[tex](2pq)^2 - 4 * (p^2 +1 )(q^2 ) <0[/tex]
[tex]4p^2 q^2 - 4p^2q^2 -4q^2<0[/tex]
[tex]-4q^2 < 0[/tex]
[tex]\mathrm{No\:Solution\:for}\:p\in \mathbb{R}[/tex]
