What are the solutions to the equation 7x^4-42x^2=35x ?

The solutions to the equation 7x^4-42x^2=35x are x=0,x=-1,x=(1+sqrt(21))/2 ,x=(1-sqrt(21))/2

Find the zeros of 7x^4-42x^2=35x

The zeros of 7x^4-42x^2=35x are x=0,x=-1,x=(1+sqrt(21))/2 ,x=(1-sqrt(21))/2

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7 x^{4} - 42 x^{2} = 35 x

x = 0, x = - 1, x = \frac{1 + 21}{2}, x = \frac{1 - 21}{2}

Decimal Notation

x = 0, x = - 1, x = 2.79128 \dots, x = - 1.79128 \dots

Solution steps

7 x^{4} - 42 x^{2} = 35 x

Move

35 x

to the left side

7 x^{4} - 42 x^{2} - 35 x = 0

Factor

7 x^{4} - 42 x^{2} - 35 x : 7 x (x + 1) (x^{2} - x - 5)

7 x (x + 1) (x^{2} - x - 5) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or x + 1 = 0 or x^{2} - x - 5 = 0

Solve

x + 1 = 0 : x = - 1

Solve

x^{2} - x - 5 = 0 : x = \frac{1 + 21}{2}, x = \frac{1 - 21}{2}

The solutions are

x = 0, x = - 1, x = \frac{1 + 21}{2}, x = \frac{1 - 21}{2}

5 x^{4} - 20 x^{3} = 0

y, y^{3} = x^{3} - x

- 4 x^{3} - 24 x^{2} - 72 x - 80 = 0

2 x^{3} - 9 x^{2} + 27 = 0

- x (x + 5) (x^{2} - 13) = 0

What are the solutions to the equation 7x^4-42x^2=35x ?
The solutions to the equation 7x^4-42x^2=35x are x=0,x=-1,x=(1+sqrt(21))/2 ,x=(1-sqrt(21))/2
Find the zeros of 7x^4-42x^2=35x
The zeros of 7x^4-42x^2=35x are x=0,x=-1,x=(1+sqrt(21))/2 ,x=(1-sqrt(21))/2