What are the solutions to the equation x^3-4x^2+7x=0 ?

The solutions to the equation x^3-4x^2+7x=0 are x=0,x=2+sqrt(3)i,x=2-sqrt(3)i

Find the zeros of x^3-4x^2+7x=0

The zeros of x^3-4x^2+7x=0 are x=0,x=2+sqrt(3)i,x=2-sqrt(3)i

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x^{3} - 4 x^{2} + 7 x = 0

x = 0, x = 2 + 3 i, x = 2 - 3 i

Solution steps

x^{3} - 4 x^{2} + 7 x = 0

Factor

x^{3} - 4 x^{2} + 7 x : x (x^{2} - 4 x + 7)

x (x^{2} - 4 x + 7) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or x^{2} - 4 x + 7 = 0

Solve

x^{2} - 4 x + 7 = 0 : x = 2 + 3 i, x = 2 - 3 i

The solutions are

x = 0, x = 2 + 3 i, x = 2 - 3 i

7 x^{5} = 35 - 29

r^{3} - 3 r^{2} + 6 r - 6 = 0

s^{3} + 6 s^{2} + 11 s + 6 = 0

(4 + 3 x) \cdot x^{5} = (4 + 3 x)^{5}

4.328 = 541 (1 + \frac{y}{4})^{15}

What are the solutions to the equation x^3-4x^2+7x=0 ?
The solutions to the equation x^3-4x^2+7x=0 are x=0,x=2+sqrt(3)i,x=2-sqrt(3)i
Find the zeros of x^3-4x^2+7x=0
The zeros of x^3-4x^2+7x=0 are x=0,x=2+sqrt(3)i,x=2-sqrt(3)i