Question

solve this ...............​

Answer 1

Answer:

c² = a²(1+m²)

Step-by-step explanation:

Given Quadratic equation :

(1+m²)x² + 2mcx+(c²- a²)=0

Compare above equation with

Ax²+Bx+C=0 ,we get

A=(1+m²), B = 2mc, C = (c²-a²)

Given, Equation has equal roots.

So, Discreminant (D) = 0

=> B²-4AC = 0

=> (2mc)²-4(1+m²)(c²-a²)=0

=> 4m²c²-4(c²-a²+m²c²-m²a²)=0

=> 4[m²c²-(c²-a²+m²c²-m²a²)]=0

=> m²c²-c²+a²-m²c²+m²a² =0

=> -c²+a²+m²a²=0

=> a²(1+m²) = c²

Therefore,

c² = a²(1+m²)

Hope it helps!

Answer 2

Question :

If ( 1+m²) x² + 2mcx + (c² - a² ) = 0 , then show that , c² = a² ( 1+m²).

Solution :

$\sf \: (1 + {m}^{2}) {x}^{2} + 2mcx + ( {c}^{2} - {a}^{2} ) = 0.............(i)$

Comparing eq(1) with ax²+bx+c = 0.

Here,

• a = (1+m²)
• b = 2mc
• c = (c²-a²)

The roots of the quadratic equation (i) are equal.

Therefore,

• b² - 4ac = 0

$\implies \sf \: {(2mc)}^{2} - 4(1 + {m}^{2})( {c}^{2} - {a}^{2} ) = 0 \\ \\ \implies \sf \: 4 {m}^{2} {c}^{2} - 4( {c}^{2} + {c}^{2} {m}^{2} - {a}^{2} - {a}^{2} {m}^{2} ) = 0 \\ \\ \implies \sf \: 4 {m}^{2} {c}^{2} - 4 {c}^{2} - 4 {c}^{2} {m}^{2} + 4 {a}^{2} + 4 {a}^{2} {m}^{2} = 0 \\ \\ \implies \sf \: 4 {c}^{2} = 4 {a}^{2} + 4 {a}^{2} {m}^{2} \\ \\ \implies \sf \: 4 {c}^{2} = 4 {a}^{2} ( 1 + {m}^{2}) \\ \\ \implies \sf \: {c}^{2} = {a}^{2} (1 + {m}^{2} )$

Hence proved !!