Question

If the quadratic equation (a2
– b2
)x2
+ (b2
– c2
)x + (c2
– a2
) = 0 has equal roots, then which of the following is

true?



(A) b2
+ c2
= a2

(B) b2 + c2
= 2a2

(C) b2
– c2
= 2a2

(D) a2
= b2 + 2c2

Answer 1

Answer :

The given quadratic equation is

( a² - b² ) x² + ( b² - c² ) x + ( c² - a² ) = 0

For equal roots, we must have

D = 0, where D is the discriminant

⇒ ( b² - c² )² - 4 ( a² - b² ) ( c² - a² ) = 0

⇒ b⁴ - 2b²c² + c⁴ - 4a²c² + 4a⁴ + 4b²c² - 4a²b² = 0

⇒ 4a⁴ + b⁴ + c⁴ - 4a²b² + 2b²c² - 4c²a² = 0

⇒ ( 2a² - b² - c² )² = 0

∴ 2a² - b² - c² = 0

⇒ b² + c² = 2a²

Thus, option ( C ) is correct.

#MarkAsBrainliest

Answer 2

Solution :

We have given a quadratic polynomial  ( a² - b² ) x² + ( b² - c² ) x + ( c² - a² ) = 0

According to the information given to us

The above quadratic polynomial has equal roots

In Condition of equal roots we know that D = 0

Where Discriminant = $b^2 - 4ac$

In order to  know which condition is true for the quadratic polynomial ( a² - b² ) x² + ( b² - c² ) x + ( c² - a² ) = 0  we need to substitute the values in the  discriminant  

As be already know with D = 0

∴ $b^2 - 4ac = 0$

Where ,

a = ( a² - b² )

b =  ( b² - c² )

c =  ( c² - a² )

⇒ ( b² - c² )² - 4 ( a² - b² ) ( c² - a² ) = 0

⇒ $( b^4 - 2b^2c^2 + c^4) - 4 ( a^2c^2 + a^4 + b^2c^2 - a^2b^2) = 0$

( ∴ $a^2 - 2ab + b^2$ )

⇒  b⁴ - 2b²c² + c⁴ - 4a²c² + 4a⁴ + 4b²c² - 4a²b² = 0

⇒ 4a⁴ + b⁴ + c⁴ - 4a²b² + 2b²c² - 4c²a² = 0

⇒ ( 2a² - b² - c² )² = 0

⇒ ∴ 2a² - b² - c² = 0

⇒ $2a^2 = b^2 + c^2$

⇒ $b^2 + c^2 = 2a^2$

Our obtained result matched with option (c) . Implies , $b^2 + c^2 = 2a^2$  is true for a quadratic polynomial ( a² - b² ) x² + ( b² - c² ) x + ( c² - a² ) = 0  has equal roots