p2 x2 +(p2_q2)x_q2=0 by qudratic formula
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Answered by
0
Step-by-step explanation:
We have p
2
x
2
+(p
2
−q
2
)x−q
2
=0
Comparing this equation to ax
2
+bx+c=0, we have
a=p
2
,b=p
2
−q
2
and c=−q
2
∴ D−D=b
2
−4ac=(p
2
−q
2
)
2
−4×p
2
+(−q)
2
=(p
2
−q
2
)
2
+4p
2
q
2
=(p
2
−q
2
)
2
>0
So, the given equation has real roots given by
α=
2a
−b+
D
=
2p
2
(p
2
−q
2
)+(p
2
+q
2
)
=
2p
2
2q
2
=
p
2
q
2
and β=
2a
−b−
D
=
2p
2
−(p
2
−q
2
)−(p
2
+q
2
)
=
2p
2
−2q
2
=−1
Hence roots are
p
2
q
2
and −1
Answered by
1
Answer:
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