Question

2. Solve the following quadratic equations using quadratic formula

Answer 1

In the given equation,



a = a²
b = (a² - b²)
c = -b²



Discriminant = b² - 4ac

=> (a² - b²)² - 4(a²)(-b²)
=> a⁴ + b⁴ - 2a²b² + 4a²b²
=> (a⁴ + b⁴ + 2a²b²)
=> (a² + b²)²




Now, applying quadratic equation,


$x = \frac{ - b( + - ) \sqrt{ discriminant} }{2a} \\ \\ \\ x = \frac{ - {a}^{2} + {b}^{2}( + - )( \sqrt{( {a}^{2} + {b}^{2} ) ^{2} } )}{2 {a}^{2} } \\ \\ \\ x = \frac{ - {a}^{2} + {b}^{2} ( + - )( {a}^{2} + {b}^{2} )}{2 {a}^{2} } \\ \\ \\ taking \: + ve \\ \\ x = \frac{ - {a}^{2} + {b}^{2} + {a}^{2} + {b}^{2} }{2 {a}^{2} } \\ \\ x = \frac{2 {b}^{2 } }{2 {a}^{2} } \\ \\ x = \frac{ {b}^{2} }{ {a}^{2} } \\ \\ \\ \\ \\ taking \: - ve \\ \\ x = \frac{ - {a}^{2} + {b}^{2} - {a}^{2} - {b}^{2} }{2 {a}^{2} }$


x = -2a²÷2a²


x = - 1



I hope this will help you

(-:

Answer 2

Compare given Quadratic equation

a²x² + ( a² - b²)x - b² = 0 with

Ax² + Bx + C = 0 we get,

A = a² , B = ( a² - b² ) , C = - b² ,



Discreminant ( D ) = B² - 4AC

= ( a² - b² )² - 4× a² ×( -b² )

= [ ( a²)² + ( b²)² - 2a²b² + 4a²b²

= ( a² )² + ( b² )² + 2a²b²

D = ( a² + b² )²

Quadratic Formula :

x = [ - B ± √D ]/2A

=> x = [ -(a²-b²) ± √(a²+b²)² ]/2a²

=> x = [ - a² + b² ± ( a² + b² ) ]/2a²

Therefore ,

i ) x = [-a²+b²+a²+b²]/2a²

x = 2b²/2a²

x = b²/a²

Or

ii ) x = ( -a² + b² - a² - b² )/2a²

x = -2a²/2a²

x = - 1

Therefore ,

x = b²/a² or x = -1

••••