Math, asked by murugadoss1972tmv, 9 months ago

Solve 9x 2 -6b 2 x-(a 4 -b 4 ) = 0

Answers

Answered by Uniquedosti00017
4

Answer:

refer to the attachment for the solution.

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Attachments:
Answered by Anonymous
10

Answer:

x = (b² + a²)/3 , (b² - a²)/3

Note:

• The possible values of unknown (variable) for which the equation is satisfied are called its solutions or roots .

• If x = a is a solution of any equation in x , then it must satisfy the given equation otherwise it's not a solution (root) of the equation.

• The discriminant of the the quadratic equation

Ax² + Bx + C = 0 , is given as ; D = B² - 4A•C

• If D > 0 then its roots are real and distinct.

• If D < 0 then its roots are imaginary.

• If D = 0 then its roots are real and equal.

Solution:

Here,

The given quadratic equation is :

9x² - 6b²x - (a⁴ - b⁴) = 0

Now,

=> 9x² - 6b²x - (a⁴ - b⁴) = 0

=> 9x² - 3•2b²x + ( -a⁴ + b⁴) = 0

=> 9x² - 3•(b² + b²)x + (b⁴ - a⁴) = 0

=> 9x² - 3•(b² + a² + b² - a²)x + (b² - a²)•(b² + a²) = 0

=> 9x² - 3(b²+a²)x - 3(b²-a²)x + (b²-a²)•(b²+a²) = 0

=> 3x•[3x - (b² + a²)] - (b² - a²)•[3x - (b² + a²)] = 0

=> [3x - (b² + a²)]•[3x - (b² - a²)] = 0

Case1

=> 3x - (b² + a²) = 0

=> 3x = b² + a²

=> x = (b² + a²)/3

Case2

=> 3x - (b² - a²) = 0

=> 3x = b² - a²

=> x = (b² - a²)/3

Hence,

The roots of the given quadratic equation are :

x = ( + )/3 , ( - )/3

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