If the quadratic equation (1+a²)b²x²+ 2abcx + (c²-m²)=0 in x has equal roots, prove that
c² = m² ( 1 + a²).
Answers
Answer:
quadratic equation (1 + a^2)b^2\text { x}^2 + 2abc \text { x }+ (c^2 - m^2) = 0(1+a
2
)b
2
x
2
+2abc x +(c
2
−m
2
)=0
To prove: c^2=m^2(1+a^2)c
2
=m
2
(1+a
2
)
Step-by-step explanation:
As we know
Ax^2+Bx+C=0Ax
2
+Bx+C=0 is the general quadratic equation
As D=B^2-4ACD=B
2
−4AC where D is discriminant
D=0 when the quadratic equation has equal roots
So we get
\begin{gathered}A= (1+a^2)b^2\\\\B=2abc\\\\C=c^2-m^2\end{gathered}
A=(1+a
2
)b
2
B=2abc
C=c
2
−m
2
So
D= (2abc)^2-4(1+a^2)b^2(c^2-m^2)D=(2abc)
2
−4(1+a
2
)b
2
(c
2
−m
2
)
Roots are equal therefore D= 0
\begin{gathered}4a^2b^2c^2-4(1+a^2)b^2(c^2-m^2)=0\\\\\Rightarrow 4b^2(a^2c^2-(1+a^2)(c^2-m^2))=0\\\\\Rightarrow (a^2c^2-c^2-a^2c^2+m^2=m^2a^2))=0\\\\\Rightarrow -c^2+m^2+m^2a^2=0\\\\\Rightarrow c^2=m^2a^2+m^2\\\\\Rightarrow c^2=m^2(1+a^2)\end{gathered}
4a
2
b
2
c
2
−4(1+a
2
)b
2
(c
2
−m
2
)=0
⇒4b
2
(a
2
c
2
−(1+a
2
)(c
2
−m
2
))=0
⇒(a
2
c
2
−c
2
−a
2
c
2
+m
2
=m
2
a
2
))=0
⇒−c
2
+m
2
+m
2
a
2
=0
⇒c
2
=m
2
a
2
+m
2
⇒c
2
=m
2
(1+a
2
)