For p, q belongs to R; the roots of (p^2+q^2) x^2 +2x (p+q) +2 =0 are
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( P² + q²)x² +2x(P +q) +2 =0
use quadratic formula,
x = { -b ±√D }/2a
D = 4(P+q)² -8(p²+q²)
=4{ -p²-q² +2pq }
=4 { -(p-q)² }
=-4(p-q)²
x ={ -2(P+q)±2√-(p-q)²}/2(p² + q²)
={ -2(P+q) ±2i(p-q)}/2(p²+q²)
use quadratic formula,
x = { -b ±√D }/2a
D = 4(P+q)² -8(p²+q²)
=4{ -p²-q² +2pq }
=4 { -(p-q)² }
=-4(p-q)²
x ={ -2(P+q)±2√-(p-q)²}/2(p² + q²)
={ -2(P+q) ±2i(p-q)}/2(p²+q²)
abhi178:
dear this question related to complex no
how
D = 2(p+q)² -2×4(p²+q²) = -4(p-q)² < 0
so, no real roots present here
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