Solve 9x 2 -6b 2 x-(a 4 -b 4 ) = 0
Answers
Answer:
refer to the attachment for the solution.
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 = (b² + a²)/3 , (b² - a²)/3