Respuesta:
q = 0
p > 0 y r < 0
o
p < 0 y r > 0
Explicación paso a paso:
px² + qx + r = 0
[tex]x'+x"=\frac{-q}{p} \\\\x'=-x"\\\\x'-x"=-\frac{q}{p}\\\\0 = -\frac{q}{p} \\\\q=0\\\\px^2+r=0\\\\p>0~y~r<0\\o\\p<0 ~y~r>0[/tex]
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Respuesta:
q = 0
p > 0 y r < 0
o
p < 0 y r > 0
Explicación paso a paso:
px² + qx + r = 0
[tex]x'+x"=\frac{-q}{p} \\\\x'=-x"\\\\x'-x"=-\frac{q}{p}\\\\0 = -\frac{q}{p} \\\\q=0\\\\px^2+r=0\\\\p>0~y~r<0\\o\\p<0 ~y~r>0[/tex]