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Answer variants:
real and unequal roots
r2+s2
4(psqr)2>0
real and equal
p2+q2
Δ=0
4(psqr)2<0
b24ac<0
4qrqr2=0
4qrqr2=0
no real root
b24ac>0
2(pr+qs)
Prove that the equation \(x^2(p^2 + q^2) + 2x(pr + qs) + r^2 + s^2 = 0\) has no real roots. If \(ps = qr\), then show that the roots are real and equal.
 
Answer:
 
Here, \(a =\)
, \(b =\)
, \(c =\)
 
\(\Delta = b^2 - 4ac =\)
 
Since \(\Delta =\)
, the given equation has
.
 
If \(ps = qr\), then \(\Delta =\)
.
 
Therefore,
 if \(ps = qr\) and so the roots will be
.
 
Hence, we proved.