Question

Is the equation in bracket 1 + m square into x square + to MCX + bracket c square minus A square equal to zero has equal roots prove that c square equal to a square in bracket oneplus m square

Answer 1

Answer:

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

⇒ b2 - 4ac = 0

⇒ (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

⇒ m2a2 - c2 + a2 = 0

⇒ a2(1 + m2) - c2 = 0  

⇒ c2  = a2(1 + m2)

Step-by-step explanation:

Answer 2

$\large{\text{\underline{SOLUTION :}}}$  

$<b>Given</b>$ : (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

$<b>D = 0 (has equal roots)</b>$

(2 mc)² - 4(1 + m²)(c² - a²) = 0

4m²c² - 4(c² - a² + m²c² - m²a²) = 0

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

$<b>4m²a² - 4c² + 4a²  = 0</b>$

4(m²a² - c² + a² ) = 0

m²a² - c² + a² = 0

a² + m²a² - c² = 0

a²(1 + m²) - c² = 0  

$<b>c²  = a²(1 + m²)</b>$

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