Question

Solve 2a2x2 + b(6a2 + 1)X + 3b2 = 0​

Answer 1

Answer:

-3b

Step-by-step explanation:

D= b²-4ac

  =b(6a²b²+b²+12a²b²-24a²t

  =36a²*²b²-12a²b²

x=-b+ -√D

 =-6a²b-b-6a²b+b/4a²

 =-12a²b/4a

 =-3b

Answer 2

Answer:

x = -3b

x = -b/2$a^{2}$

Step-by-step explanation:

2$a^{2}$$x^{2}$ + b(6

2$a^{2}$$x^{2}$ +(6

divide the whole term by 2$a^{2}$

we get ; $x^{2}$ + (6$a^{2}$b +b/2

             $x^{2}$ + (3b +b/2$a^{2}$ )x + 3b*b/2

             $x^{2}$ - (-3b -b/2$a^{2}$ )x + 3b*b/2  ________(1)

we know if alpha and beta r roots of a quadratic equation . then, it can be represented as;    $x^{2}$ - (alpha +beta)x +alpha*beta

now relate it with __(1)

we get

alpha = -3b , beta = -b/2$a^{2}$

x = -3b , x = -b/2$a^{2}$

hope it helps

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