Solve (-k)^2-4(1)(4)=0

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

Solve for k

$$k=-4$$

$$k=4$$

Steps Using Difference of Squares

Calculate -k to the power of 2 and get k^{2}.

$$k^{2}-4\times 1\times 4=0$$

Multiply 4 and 1 to get 4.

$$k^{2}-4\times 4=0$$

Multiply 4 and 4 to get 16.

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

Consider k^{2}-16. Rewrite k^{2}-16 as k^{2}-4^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(k-4\right)\left(k+4\right)=0$$

To find equation solutions, solve k-4=0 and k+4=0.

$$k=4 k=-4$$

Steps by Finding Square Root

Calculate -k to the power of 2 and get k^{2}.

$$k^{2}-4\times 1\times 4=0$$

Multiply 4 and 1 to get 4.

$$k^{2}-4\times 4=0$$

Multiply 4 and 4 to get 16.

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

Add 16 to both sides. Anything plus zero gives itself.

$$k^{2}=16$$

Take the square root of both sides of the equation.

$$k=4 k=-4$$

Steps Using the Quadratic Formula

Calculate -k to the power of 2 and get k^{2}.

$$k^{2}-4\times 1\times 4=0$$

Multiply 4 and 1 to get 4.

$$k^{2}-4\times 4=0$$

Multiply 4 and 4 to get 16.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{0±\sqrt{0^{2}-4\left(-16\right)}}{2}$$

Square 0.

$$k=\frac{0±\sqrt{-4\left(-16\right)}}{2}$$

Multiply -4 times -16.

$$k=\frac{0±\sqrt{64}}{2}$$

Take the square root of 64.

$$k=\frac{0±8}{2}$$

Now solve the equation k=\frac{0±8}{2} when ± is plus. Divide 8 by 2.

$$k=4$$

Now solve the equation k=\frac{0±8}{2} when ± is minus. Divide -8 by 2.

$$k=-4$$

The equation is now solved.

$$k=4 k=-4$$