Question

lf the equation (1+m²)x²+2mcx +(c²-a²)= 0 has equal roots ,then prove c²= a²(1+m²).

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Answer 1

$\large{\mathbf{HELLO \: \: DEAR,}}$

$\mathbf{HERE}, \\ \\ \mathit{(1+m^2)x^2+2mcx +(c^2-a^2)= 0 }\\ \\ \mathit{WHERE,} \\ \\ \mathit{a = (1 + m^2) , b = 2mc , c = (c^2 - a^2)} \\ \\ \mathbf{we \:\:know \:\:that:-}\\ \\ \mathbf{ EQUAL \: \: ROOTS \: \: DISCRIMINANT = 0} \\ \\ \mathbf{D = b^2 - 4ac = 0} \\ \\ \mathit{D = (2mc)^2 - 4[(1 + m^2)*(c^2 - a^2)]}\\ \\ \mathit{D= 4(mc)^2 - 4[c^2 - a^2 + (mc)^2 - (ma)^2]} \\ \\ \mathit{D = 4(mc)^2 - 4c^2 + 4a^2 - 4(mc)^2 + 4(ma)^2} \\ \\ \mathit{D = -4c^2 + 4a^2 + 4(ma)^2} \\ \\ \mathit{0 = 4(ma)^2 + 4a^2 - 4c^2}\\ \\ \to\to\to\to\to\to \therefore \boxed{D = 0} \\ \\ \mathit{4c^2 = 4a^2(m^2 + 1)} \\ \\ \mathit{c^2 = a^2(1 + m^2)}$

$\large{\mathbf{\underline{I \: \: HOPE \: \: ITS \: \: HELP \: \: YOU \: \: DEAR, \: \: THANKS}}}$

Answer 2

Given Equation is (1 + m^2)x^2 + 2mcx + (c^2 - a^2) = 0.

On comparing with ax^2 + bx + c = 0,

we get a = (1 + m^2), b = 2mc, c = (c^2 - a^2).

Given that the equation has real roots.

D = > b^2 - 4ac = 0

= > (2mc)^2 - 4(1 + m^2)(c^2 - a^2) = 0

= > 4m^2c^2 - 4(c^2 - a^2 + m^2c^2 - m^2a^2) = 0

= > 4m^2c^2 - 4c^2 + 4a^2 - 4m^2c^2 + 4m^2a^2 = 0

= > -4c^2 + 4a^2 + 4m^2a^2 = 0

= > 4c^2 - 4a^2 - 4m^2a^2 = 0

= > 4c^2 = 4a^2 + 4m^2a^2

= > 4c^2 = 4(a^2 + m^2a^2)

= > c^2 = a^2(1 + m^2).


Hope this helps!