What are the solutions to the equation solvefor x,K(x)^2=4(128x+x^2) ?

The solutions to the equation solvefor x,K(x)^2=4(128x+x^2) are x=0,x=(512)/(K-4);\quad K\ne 4

Find the zeros of solvefor x,K(x)^2=4(128x+x^2)

The zeros of solvefor x,K(x)^2=4(128x+x^2) are x=0,x=(512)/(K-4);\quad K\ne 4

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solve for

x, K (x)^{2} = 4 (128 x + x^{2})

x = 0, x = \frac{512}{K - 4}; K \neq = 4

Solution steps

K x^{2} = 4 (128 x + x^{2})

Move

4 (128 x + x^{2})

to the left side

K x^{2} - 4 (128 x + x^{2}) = 0

Factor

K x^{2} - 4 (128 x + x^{2}) : x (x K - 4 (128 + x))

x (x K - 4 (128 + x)) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or x K - 4 (128 + x) = 0

Solve

x K - 4 (128 + x) = 0 : x = \frac{512}{K - 4}; K \neq = 4

The solutions to the quadratic equation are:

x = 0, x = \frac{512}{K - 4}; K \neq = 4

\frac{( x - 1 ) ^{2}}{9} - \frac{( y - 5 ) ^{2}}{16} = 1

9 x^{2} - 16 y^{2} - 18 x - 64 y - 200 = 0

4 x^{2} - 16 y^{2} - 64 y - 128 = 0

x^{2} + 5^{2} = y^{2}

x^{2} - 9 y^{2} + 2 x - 5 yy - 80 = 0

What are the solutions to the equation solvefor x,K(x)^2=4(128x+x^2) ?
The solutions to the equation solvefor x,K(x)^2=4(128x+x^2) are x=0,x=(512)/(K-4);\quad K\ne 4
Find the zeros of solvefor x,K(x)^2=4(128x+x^2)
The zeros of solvefor x,K(x)^2=4(128x+x^2) are x=0,x=(512)/(K-4);\quad K\ne 4