Question

solve this using quadriatic formula
4x^2-2(a^2+b^2)x+a^2b^2=0.

Answer 1

Answer:

x= a²/2 or b²/2

Step-by-step explanation:

Given Quadratic equation :

4x² -2(a²+b²)x + a²b² = 0

Comparing it to a general quadratic equation ax²+bx+c=0,

we get a = 4 , b= -2(a²+b²), c= a²b²

By quadratic formula:

x = (-b ± √(b²-4ac))/2a

Putting the corresponding values, we get:

x = (2(a²+b²) ±√(4(a²+b²)²-4(4)(a²b²)))/2(4)

or x = (2a²+2b² ± 2√((a²+b²)²-4a²b²))/8    {Since 4 is common in both factors,  

                                                                     we can take it outside sq.root}

or x = (a²+b² ± √(a²-b²)²)/4         {I hope u know that : (a²+b²)²-4a²b² =(a²-b²)²}

or x = (a² + b² ± (a²-b²))/4

 Thus,

x = (a²+b² + a² - b²)/4      or  x = (a²+b²-a²+b²)/4

x= 2a²/4  or x = 2b²/4

Thus, x = a²/2 or b²/2

Answer 2

Answer:

a²/2, b²/2

Step-by-step explanation:

Given Equation is 4x² - 2(a² + b²)x + a² b² = 0

Here, a = 4, b = -2(a² + b²), c = a² b²

∴ D =  b² - 4ac

⇒ [-2(a² + b²)]² - 4(4)(a²b²)

⇒ 4(a² + b²)² - 16(a²b²)

⇒ 4(a⁴ + b⁴ + 2a²b²) - 16a²b²

⇒ 4a⁴ + 4b⁴ + 8a²b² - 16a²b²

⇒ (2a² - 2b²)²

Now,

⇒ x = (-b ± √D)/2a

      = [2(a² + b²) ± √(2a² - 2b²)²]/8

      = [2(a² + b²) ± (2a² - 2b²)]/8

(i)

2(a² + b²) + (2a² - 2b²)/8

= 2a² + 2b² + 2a² - 2b²/8

= 4a²/8

= a²/2.

(ii)

2(a² + b²) - (2a² - 2b²)/8

= (2a² + 2b² - 2a² + 2b²)/8

= 4b²/8

= b²/2.

Therefore, the roots are a²/2, b²/2.

Hope it helps!