Question

3y в квадрате+8у+4=0
-6х в квадрате+13х-6=0
х в четвертом степене-13х в квадрате+36=0
Помогите решить пожалуйста..

Answer 1

3Y^2 + 8Y + 4 = 0
D = 64 - 4*3*4 = 64 - 48 = 16 
V D = 4
X1 = ( - 8 + 4 ) : 6 = ( - 4/6 ) = ( - 2/3 )
X2 = ( - 12 ) : 6 = ( - 2 )
===========================
-6X^2 + 13X - 6 = 0 
D = 169 - 4*(-6)*(-6) = 169 - 144 = 25
V D = 5
X1 = ( - 13 + 5 ) : ( - 12 ) = + 8/12 = 4/6 = 2/3 
X2 = ( - 18 ) : ( - 12 ) = + 1.5 
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X^4 - 13X^2 + 36 = 0
X^2 = A ( A > 0 )
A^2 - 13A + 36 = 0
D = 169 - 4*1*36 = 169 - 144 = 25
V D = 5
A1 = ( - 1 + 5 ) : 2 = 2 
A2 = ( - 6 ) : 2 = ( - 3 ) ( < 0 )
X^2 = 2 ---> X = V 2 
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Answer 2

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