(x^2-3x)^2+17(x^2-3x)=-18-3(x^2-3x)^2 solve equations... be deeply grateful How to calculate x when X & sup2; - 3x = - 2 and equal to - 9 / 4?

(x^2-3x)^2+17(x^2-3x)=-18-3(x^2-3x)^2 solve equations... be deeply grateful How to calculate x when X & sup2; - 3x = - 2 and equal to - 9 / 4?

(x ^ 2-3x) ^ 2 + 17 (x ^ 2-3x) = - 18-3 (x ^ 2-3x) ^ 24 (x ^ 2-3x) ^ 2 + 17 (x ^ 2-3x) + 18 = 0, then x ^ 2-3x as a whole y can be reduced to 4Y ^ 2 + 17Y + 18 = 0 (y + 2) (4Y + 9) = 0, y = - 2, y = - 9 / 4x ^ 2-3x = y when y = - 2, x = 1 or 2 when y = - 9 / 4, x = 3 / 2 x ^ 2-3x

When x is a positive integer_____ When x + 17 > 20 and 3x-5

∵x+17>20∴x>3
∵3x-5≤10∴x≤5
∴3﹤x≤5
A positive integer x = 4,5

41-3x = 17 method of solving equation

Move x to the right and 17 to the left
41-17=3x
24=3x
Divide by 3
8=x
x=8

X + (1 / x) = 3, find the value of x ^ 4 + 3x ^ 3-16x ^ 2 + 3x-17
x+(1/x)=3
Find the value of x ^ 4 + 3x ^ 3-16x ^ 2 + 3x-17

x+1/x=3
x^2+1=3x
x^2=3x-1
x^4+3x^3-16x^2+3x-17
=x^2(3x-1)+3x^3-16x^2+3x-17
=6x^3-17x^2+3x-17
=6x(3x-1)-17x^2+3x-17
=x^2-3x-17
=3x-1-3x-17
=-18