Plz answer plz plsssss faassttt answer with explaination
Answers
Answer:
c2
=a
2
(1+m
2
) if the equation \bold{(1+m^{2}) \times x^{2}+2 m c x+(c^{2}-a^{2})=0}(1+m
2
)×x
2
+2mcx+(c
2
−a
2
)=0 has equal roots.
Given:
The equation (1+m^{2}) \times x^{2}+2 m c x+(c^{2}-a^{2})=0(1+m
2
)×x
2
+2mcx+(c
2
−a
2
)=0 has equal roots.
To prove:
Prove that c^{2}=a^{2}(1+m^{2})c
2
=a
2
(1+m
2
)
Proof:
Any quadratic equation a x^{2}+b x+c=0ax
2
+bx+c=0 has real roots when the discriminant is equal to 0. That is
b^{2}-4 a c=0b
2
−4ac=0
Consider the given equation (1+m^{2}) \times x^{2}+2 m c x+(c^{2}-a^{2})=0(1+m
2
)×x
2
+2mcx+(c
2
−a
2
)=0
This equation is a quadratic equation in x.
Here b=2 m c ; a=1+m^{2} ; c=c^{2}-a^{2}b=2mc;a=1+m
2
;c=c
2
−a
2
So, b^{2}-4 a c=(2 m c)^{2}-4(1+m^{2})(c^{2}-a^{2})=0b
2
−4ac=(2mc)
2
−4(1+m
2
)(c
2
−a
2
)=0
\begin{gathered}\begin{array}{l}{\Rightarrow 4 m^{2} \times c^{2}-4(c^{2}+m^{2} \times c^{2}-a^{2}-m^{2} \times a^{2})=0} \\ {\Rightarrow m^{2} \times c^{2}-c^{2}-m^{2} \times c^{2}+a^{2}+m^{2} \times a^{2}=0} \\ {\Rightarrow c^{2}-a^{2}=m^{2} \times a^{2}}\end{array}\end{gathered}
⇒4m
2
×c
2
−4(c
2
+m
2
×c
2
−a
2
−m
2
×a
2
)=0
⇒m
2
×c
2
−c
2
−m
2
×c
2
+a
2
+m
2
×a
2
=0
⇒c
2
−a
2
=m
2
×a
2
\Rightarrow c^{2}=a^{2}(1+m^{2})⇒c
2
=a
2
(1+m
2
)
Hence proved.
\bold{c^{2}=a^{2}(1+m^{2})}c
2
=a
2
(1+m
2
) if the equation \bold{(1+m^{2}) \times x^{2}+2 m c x+(c^{2}-a^{2})=0}(1+m
2
)×x
2
+2mcx+(c
2
−a
2
)=0 has equal roots.
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