Question

If
(a +b²) ² - 2b (a + c) a + b + c=o have equal roots
prove that a,b,c are in G.P.​

Answer 1

Correct Question:

If ( a² + b² ) x² - 2b ( a + c ) x + ( b² + c² ) = 0 has equal roots, prove that a, b, c are in G.P.

Answer:

a, b, c are in G.P.

Step-by-step-explanation:

The given quadratic equation is

( a² + b² ) x² - 2b ( a + c ) x + ( b² + c² ) = 0

Comparing with ax² + bx + c = 0, we get,

• a = ( a² + b² )
• b = - 2b ( a + c )
• c = ( b² + c² )

For equal roots,

b² - 4ac = 0

∴ [ - 2b ( a + c ) ]² - [ 4 * ( a² + b² ) * ( b² + c² ) ] = 0

⇒ ( - 2ab - 2bc )² - [ ( 4a² + 4b² ) * ( b² + c² ) ] = 0

⇒ ( 4a²b² - 2 * ( - 2ab ) * ( 2bc ) + 4b²c² ) - ( 4a²b² + 4a²c² + 4b⁴ + 4b²c² ) = 0

⇒ 4a²b² + 8ab²c + 4b²c² - 4a²b² - 4a²c² - 4b⁴ - 4b²c² = 0

⇒ 4a²b² - 4a²b² + 4b²c² - 4b²c² + 8ab²c - 4a²c² - 4b⁴ = 0

⇒ 8ab²c - 4a²c² - 4b⁴ = 0

⇒ 4 ( 2ab²c - a²c² - b⁴ ) = 0

⇒ 2ab²c - a²c² - b⁴ = 0

⇒ - ( b⁴ - 2ab²c + a²c² ) = 0

⇒ b⁴ - 2ab²c + a²c² = 0

⇒ ( b² )² - 2 * ( b² ) * ac + ( ac )² = 0

⇒ ( b² - ac )² = 0

⇒ b² - ac = 0

⇒ b² = ac

⇒ b * b = a * c

⇒ b / a = c / b

∴ a, b, c are in G.P.

Hence proved!

Answer 2

Explanation:

(a-b)x² + (b-c) x+ (c - a) = 0

G.P 2a = b + c

B² – 4AC = 0

(b-c)² – [4(a-b) (c - a)] = 0

b²-2bc + c² – [4(ac-a² – bc + ab)] = 0

=> b²-2bc + c² – 4ac + 4a² + 4bc - 4ab = 0

=> b²+ 2bc + c² + 4a² – 4ac – 4ab= 0

=> (b + c - 2a)² = 0

=> b + c = 2a