7. If the equation is
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Hi ..mate!!,
Here is your answer!!!...
...we know that if euqations ax²+ bx+c= 0 has real and euqal roots than b²-4ac= 0........
similarly apply this theorem in given equation...
......see picture!!!...
hope you understand!!
Regards Brainly Star Community
#shubhendu
Here is your answer!!!...
...we know that if euqations ax²+ bx+c= 0 has real and euqal roots than b²-4ac= 0........
similarly apply this theorem in given equation...
......see picture!!!...
hope you understand!!
Regards Brainly Star Community
#shubhendu
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Solution :
Compare given Quadratic equation
( 1+m²)x² + 2mcx + c² - a² = 0 with
Ax² + Bx + C = 0 ,we get
A = 1 + m² ,
B = 2mc ,
C = c² - a² ,
Discreminant = 0
B² - 4AC = 0 [ given equal roots ]
( 2mc )² - 4( 1 + m² )(c² - a² ) = 0
=> 4m²c² -4 (c²-a² +m²c² - m²a²) = 0
=>4[ m²c² - c² + a² - m²c² + m²a²]= 0
=> -c² + a² + m²a² = 0
=> a²( 1 + m² ) = c²
Therefore ,
c² = a²( 1 + m² )
••••
Compare given Quadratic equation
( 1+m²)x² + 2mcx + c² - a² = 0 with
Ax² + Bx + C = 0 ,we get
A = 1 + m² ,
B = 2mc ,
C = c² - a² ,
Discreminant = 0
B² - 4AC = 0 [ given equal roots ]
( 2mc )² - 4( 1 + m² )(c² - a² ) = 0
=> 4m²c² -4 (c²-a² +m²c² - m²a²) = 0
=>4[ m²c² - c² + a² - m²c² + m²a²]= 0
=> -c² + a² + m²a² = 0
=> a²( 1 + m² ) = c²
Therefore ,
c² = a²( 1 + m² )
••••
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