[1+msq]xsq+[2mc]x+[csq-asq]=0 if this eqution has equal roots ,then prove that csq=asq[1+msq] please answer this
siddhartharao77:
is it (1 + m^2)x^2 + (2mcx) + (c^2 - a^2) = 0?
cant say
Its my mistake. I don't have brain to understand this question.
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Answered by
4
Given equation is:
(1 + m 2) x 2 + 2mcx + (c 2 – a 2) = 0
To prove: c 2 = a 2 (1 + m 2)
Proof: It is being given that equation has equal roots, therefore
D = b 2 – 4ac = 0 ... (1)
From the above equation, we have
a = (1 + m 2)
b = 2mc
and c= (c 2 – a 2)
Putting values of a, b and c in (1), we get
D = (2mc)2 – 4 (1 + m 2) (c 2 – a 2) = 0
⇒ 4m 2 c 2 – 4 (c 2 + c 2 m 2 – a 2 – a 2 m 2) = 0
⇒ 4m 2 c 2 – 4c 2 – 4c 2 m 2 + 4a 2 + 4a 2 m 2= 0
⇒ – 4c 2 + 4a 2 + 4a 2 m 2 = 0
⇒ 4c 2 = 4a 2 + 4a 2 m 2
⇒ c 2 = a 2 + a 2 m 2
⇒ c 2 = a 2 (1 + m 2)
[Hence proved]
(1 + m 2) x 2 + 2mcx + (c 2 – a 2) = 0
To prove: c 2 = a 2 (1 + m 2)
Proof: It is being given that equation has equal roots, therefore
D = b 2 – 4ac = 0 ... (1)
From the above equation, we have
a = (1 + m 2)
b = 2mc
and c= (c 2 – a 2)
Putting values of a, b and c in (1), we get
D = (2mc)2 – 4 (1 + m 2) (c 2 – a 2) = 0
⇒ 4m 2 c 2 – 4 (c 2 + c 2 m 2 – a 2 – a 2 m 2) = 0
⇒ 4m 2 c 2 – 4c 2 – 4c 2 m 2 + 4a 2 + 4a 2 m 2= 0
⇒ – 4c 2 + 4a 2 + 4a 2 m 2 = 0
⇒ 4c 2 = 4a 2 + 4a 2 m 2
⇒ c 2 = a 2 + a 2 m 2
⇒ c 2 = a 2 (1 + m 2)
[Hence proved]
Answered by
3
(1 + m 2) x 2 + 2mcx + (c 2 – a 2) = 0
To prove: c 2 = a 2 (1 + m 2)
Proof: It is being given that equation has equal roots, therefore
D = b 2 – 4ac = 0 ... (1)
From the above equation, we have
a = (1 + m 2)
b = 2mc
and c= (c 2 – a 2)
Putting values of a, b and c in (1), we get
D = (2mc)2 – 4 (1 + m 2) (c 2 – a 2) = 0
⇒ 4m 2 c 2 – 4 (c 2 + c 2 m 2 – a 2 – a 2 m 2) = 0
⇒ 4m 2 c 2 – 4c 2 – 4c 2 m 2 + 4a 2 + 4a 2 m 2= 0
⇒ – 4c 2 + 4a 2 + 4a 2 m 2 = 0
⇒ 4c 2 = 4a 2 + 4a 2 m 2
⇒ c 2 = a 2 + a 2 m 2
⇒ c 2 = a 2 (1 + m 2)
[Hence proved]
If it helps u PLZ MARK AS BRAINLIEST
To prove: c 2 = a 2 (1 + m 2)
Proof: It is being given that equation has equal roots, therefore
D = b 2 – 4ac = 0 ... (1)
From the above equation, we have
a = (1 + m 2)
b = 2mc
and c= (c 2 – a 2)
Putting values of a, b and c in (1), we get
D = (2mc)2 – 4 (1 + m 2) (c 2 – a 2) = 0
⇒ 4m 2 c 2 – 4 (c 2 + c 2 m 2 – a 2 – a 2 m 2) = 0
⇒ 4m 2 c 2 – 4c 2 – 4c 2 m 2 + 4a 2 + 4a 2 m 2= 0
⇒ – 4c 2 + 4a 2 + 4a 2 m 2 = 0
⇒ 4c 2 = 4a 2 + 4a 2 m 2
⇒ c 2 = a 2 + a 2 m 2
⇒ c 2 = a 2 (1 + m 2)
[Hence proved]
If it helps u PLZ MARK AS BRAINLIEST
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