Solve (4k-8)^2-4(1-k)(4)>0

$$( 4 k - 8 ) ^ { 2 } - 4 ( 1 - k ) ( 4 ) > 0$$

Solve for k

$$k\in \mathrm{R}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4k-8\right)^{2}.

$$16k^{2}-64k+64-4\left(1-k\right)\times 4>0$$

Multiply 4 and 4 to get 16.

$$16k^{2}-64k+64-16\left(1-k\right)>0$$

Use the distributive property to multiply -16 by 1-k.

$$16k^{2}-64k+64-16+16k>0$$

Subtract 16 from 64 to get 48.

$$16k^{2}-64k+48+16k>0$$

Combine -64k and 16k to get -48k.

$$16k^{2}-48k+48>0$$

To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$16k^{2}-48k+48=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 16 for a, -48 for b, and 48 for c in the quadratic formula.

$$k=\frac{-\left(-48\right)±\sqrt{\left(-48\right)^{2}-4\times 16\times 48}}{2\times 16}$$

Do the calculations.

$$k=\frac{48±\sqrt{-768}}{32}$$

Since the square root of a negative number is not defined in the real field, there are no solutions. Expression 16k^{2}-48k+48 has the same sign for any k. To determine the sign, calculate the value of the expression for k=0.

$$16\times 0^{2}-48\times 0+48=48$$

The value of the expression 16k^{2}-48k+48 is always positive. Inequality holds for k\in \mathrm{R}.

$$k\in \mathrm{R}$$