Question

ɪғ ᴛʜᴇ ᴇǫᴜᴀᴛɪᴏɴ
$(1 + {m}^{2} ) {x}^{2} \: + 2mc \: + ( {c}^{2} - {a}^{2} ) = 0$ʜᴀs ᴀ ᴇǫᴜᴀʟ ʀᴏᴏᴛs ᴘʀᴇ ᴛʜᴀᴛ
${c}^{2} = {m}^{2}$




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Answer 1

$\huge \bold \blue{Answer࿐}$

SOLUTION :  

Given : (1 + m²)x² + 2 mcx + c² - a² = 0 has equal roots

On comparing the given equation with,  ax² +  bx + c = 0

Here, a =  (1 + m²) , b = 2mc , c =  c² - a²

Discriminant , D = b² -  4ac

D = 0 (has equal roots)

(2 mc)² - 4(1 + m²)(c² - a²) = 0

4m²c² - 4(c² - a² + m²c² - m²a²) = 0

4m²c² - 4c²  + 4a² - 4m²c² + 4m²a² = 0

4m²a² - 4c² + 4a²  = 0

4(m²a² - c² + a² ) = 0

m²a² - c² + a² = 0

a² + m²a² - c² = 0

a²(1 + m²) - c² = 0  

c²  = a²(1 + m²)

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Answer 2

Answer:

Given- (1+m)x² +mcx + (c²-a²)= 0

To prove- c² = a² (1+m²)

Solution- (1 + m²) x²+2 mcx + (c²-a²)=0

we know that the condition for equal root

Determinant = 0

→ Determinant = b² - 4ac=0

Here a = 1 + m² , b= 2mc , c= c²-a²

since (2mc)² - 4 (1+ m²) ( c²-a²) = 0

→ 4c²m² -4c² - 4c² m² + 4a²+ 4a²m² = 0

→ -4c²= -4a²-4a²m²

→ 4c²= 4a²+4a²m²

→ 4c²= 4a²(1+ m²)

→ c²=a²(1+ m²)

hence proved, c²=a²(1+ m²)