(x^2-2x)^2+3x^2-2x)-4=0?

Question

Please show me the method and fomula...

Answer 1

(x^2-2x)^2+3(x^2-2x)-4=0?

Probably easiest to replace x^2 - 2x with u to get

u^2 + 3u - 4 = 0

(u+4)(u-1) = 0

u = -4 or u = 1

Then put back the x^2 - 2x for u and solve those simpler equations

Answer 2

I assume the problem reads like this because you're missing a parenthesis:
(x^2 - 2x)^2 + 3(x^2 - 2x) - 4 = 0, solve for x
Try substitution
m = x^2 - 2x
So
m^2 + 3m - 4 = 0
(m + 4)(m - 1) = 0
So either m + 4 = 0, or m - 1 = 0
m = -4, 1
If m = -4, then
x^2 - 2x = -4
x^2 - 2x + 4 = 0
referring to the quadratic equation, since b^2 - 4ac = 4 - 16 = -12 < 0, there is no real solution for m = -4. (We would have to take the square root of -12 to do so.) We disregard this possibility.
Now try m = 1
x^2 - 2x = 1
x^2 - 2x - 1 = 0
again using the quadratic equation,
x = 1 +/- sqrt(2)

Hope this helps.

Answer 3

i know 3+1=4

Answer 4

((x^2-2x)^2+3x^2-2x)-4 = 0
x^4-4x^3+4x^2+3x^2-2x-4 = 0
x^4 - 4x^3 + 7x^2 - 2x - 4 = 0

Answer 5

You are missing(or have too many) parentheses?? Very important in math.

Answer 6

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