Solve -x^2+8x=-4(x-4)^2+k

$$- x ^ { 2 } + 8 x = - 4 ( x - 4 ) ^ { 2 } + k$$

Solve for k

$$k=3x^{2}-24x+64$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$-x^{2}+8x=-4\left(x^{2}-8x+16\right)+k$$

Use the distributive property to multiply -4 by x^{2}-8x+16.

$$-x^{2}+8x=-4x^{2}+32x-64+k$$

Swap sides so that all variable terms are on the left hand side.

$$-4x^{2}+32x-64+k=-x^{2}+8x$$

Add 4x^{2} to both sides.

$$32x-64+k=-x^{2}+8x+4x^{2}$$

Subtract 32x from both sides.

$$-64+k=-x^{2}+8x+4x^{2}-32x$$

Combine 8x and -32x to get -24x.

$$-64+k=-x^{2}-24x+4x^{2}$$

Add 64 to both sides.

$$k=-x^{2}-24x+4x^{2}+64$$

Combine -x^{2} and 4x^{2} to get 3x^{2}.

$$k=3x^{2}-24x+64$$

Solve for x (complex solution)

$$x=\frac{\sqrt{3k-48}}{3}+4$$

$$x=-\frac{\sqrt{3k-48}}{3}+4$$

Solve for x

$$x=\frac{\sqrt{3k-48}}{3}+4$$

$$x=-\frac{\sqrt{3k-48}}{3}+4\text{, }k\geq 16$$