(i) Prove that the roots of the equation x^2 - 2ax + a^2 - b^2 - c^2 = 0 are always real, a, b, c E R.
(ii) Show that roots of the equation (x - a) (x - b) = abx^2; a, b E R are always real. When are they equal?
Answers
Answer:
(i). Given Quadratic Equation,
x² - 2ax + a² - b² - c² = 0
a , b , c ∈ R
To prove: Roots of the Equation are always real.
If the standard Quadratic Equation, Ax² + Bx + C = 0
Then Discriminant is given by, D = B² - 4AC
From given Equation,
A = 1 , B = -2a , C = a² - b² - c²
D = (-2a)² - 4(1)(a² - b² - c²) = 4a² - 4a² + b² + c²
= b² + c²
Since D > 0
⇒ Roots are Real and Distinct.
Hence Proved.
(ii). Given Quadratic Equation,
( x - a )( x - b ) = abx²
a , b , c ∈ R
To prove: Roots of the Equation are always real.
If the standard Quadratic Equation, Ax² + Bx + C = 0
Then Discriminant is given by, D = B² - 4AC
Consider,
( x - a )( x - b ) = abx²
x² - ( a + b )x + ab = abx²
( 1 - ab )x² - ( a + b )x + ab = 0
From given Equation,
A = 1 - ab , B = - ( a + b ) , C = ab
D = ( -a - b )² - 4(1-ab)(ab) = a² + b² + 2ab - 4ab + 4a²b²
= a² + b² + 4a²b² - 4ab
Since D > 0
⇒ Roots are Real and Distinct.
Hence Proved.
Roots are equal.
⇒ D = 0
a² + b² + 4a²b² - 4ab = 0
⇒ a = 0 and b = 0
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