What are the solutions to the equation 8(1.2-x^2)^2=0.5x^2 ?

The solutions to the equation 8(1.2-x^2)^2=0.5x^2 are

Find the zeros of 8(1.2-x^2)^2=0.5x^2

The zeros of 8(1.2-x^2)^2=0.5x^2 are

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8 (1.2 - x^{2})^{2} = 0.5 x^{2}

x = \frac{197 + 1945}{4 10}, x = - \frac{197 + 1945}{4 10}, x = \frac{197 - 1945}{4 10}, x = - \frac{197 - 1945}{4 10}

Solution steps

8 (1.2 - x^{2})^{2} = 0.5 x^{2}

Expand

8 (1.2 - x^{2})^{2} : 11.52 - 19.2 x^{2} + 8 x^{4}

11.52 - 19.2 x^{2} + 8 x^{4} = 0.5 x^{2}

Move

0.5 x^{2}

to the left side

8 x^{4} - 19.7 x^{2} + 11.52 = 0

Rewrite the equation with

u = x^{2}

and

u^{2} = x^{4}

8 u^{2} - 19.7 u + 11.52 = 0

Solve

8 u^{2} - 19.7 u + 11.52 = 0 : u = \frac{197 + 1945}{160}, u = \frac{197 - 1945}{160}

u = \frac{197 + 1945}{160}, u = \frac{197 - 1945}{160}

Substitute back

u = x^{2},

solve for

x

Solve

x^{2} = \frac{197 + 1945}{160} : x = \frac{197 + 1945}{4 10}, x = - \frac{197 + 1945}{4 10}

Solve

x^{2} = \frac{197 - 1945}{160} : x = \frac{197 - 1945}{4 10}, x = - \frac{197 - 1945}{4 10}

The solutions are

x = \frac{197 + 1945}{4 10}, x = - \frac{197 + 1945}{4 10}, x = \frac{197 - 1945}{4 10}, x = - \frac{197 - 1945}{4 10}

76 = \frac{1}{2} (h (h^{2} + 3))

x^{2} = 8 x^{15}

4 x^{3} - 6 x^{2} - 20 x + 30 = 0

(x^{2} - 8 x + 15) (x + 1) = 0

x^{4} - 4 x^{3} + 7 x^{2} - 8 x + 10 = 0

What are the solutions to the equation 8(1.2-x^2)^2=0.5x^2 ?
The solutions to the equation 8(1.2-x^2)^2=0.5x^2 are
Find the zeros of 8(1.2-x^2)^2=0.5x^2
The zeros of 8(1.2-x^2)^2=0.5x^2 are