somtochikwendu somtochikwendu
  • 14-11-2019
  • Mathematics
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given that n^2+2pn+p^2+q^2=r^2 has real roots show that r^2=q^2

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Аноним Аноним
  • 14-11-2019

It doesn't seem true, and here's a counterexample: observe that the first three terms form a perfect square. You can rewrite the equation as

[tex](n+p)^2+q^2=r^2[/tex]

This is basically the Pythagorean theorem applied to a triangle with sides n+p, q and r. For example, pick:

[tex]n=1, p=2, q=4, r^5[/tex]

The expression becomes

[tex]1^2+2\cdot 2\cdot 1+2^2+4^2=5^2 \iff 1+4+4+16=25 \iff 25=25[/tex]

Which is true, even if

[tex]r^2\neq q^2[/tex]

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