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

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

Solve for k

$$k = \frac{\sqrt{5}}{2} \approx 1.118033989$$

$$k = -\frac{\sqrt{5}}{2} \approx -1.118033989$$

Steps by Finding Square Root

Use the distributive property to multiply -2 by 2k-1.

$$4-4k+2=\left(2k-1\right)^{2}$$

Add 4 and 2 to get 6.

$$6-4k=\left(2k-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2k-1\right)^{2}.

$$6-4k=4k^{2}-4k+1$$

Subtract 4k^{2} from both sides.

$$6-4k-4k^{2}=-4k+1$$

Add 4k to both sides.

$$6-4k-4k^{2}+4k=1$$

Combine -4k and 4k to get 0.

$$6-4k^{2}=1$$

Subtract 6 from both sides.

$$-4k^{2}=1-6$$

Subtract 6 from 1 to get -5.

$$-4k^{2}=-5$$

Divide both sides by -4.

$$k^{2}=\frac{-5}{-4}$$

Fraction \frac{-5}{-4} can be simplified to \frac{5}{4} by removing the negative sign from both the numerator and the denominator.

$$k^{2}=\frac{5}{4}$$

Take the square root of both sides of the equation.

$$k=\frac{\sqrt{5}}{2} k=-\frac{\sqrt{5}}{2}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply -2 by 2k-1.

$$4-4k+2=\left(2k-1\right)^{2}$$

Add 4 and 2 to get 6.

$$6-4k=\left(2k-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2k-1\right)^{2}.

$$6-4k=4k^{2}-4k+1$$

Subtract 4k^{2} from both sides.

$$6-4k-4k^{2}=-4k+1$$

Add 4k to both sides.

$$6-4k-4k^{2}+4k=1$$

Combine -4k and 4k to get 0.

$$6-4k^{2}=1$$

Subtract 1 from both sides.

$$6-4k^{2}-1=0$$

Subtract 1 from 6 to get 5.

$$5-4k^{2}=0$$

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$-4k^{2}+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 0 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{0±\sqrt{0^{2}-4\left(-4\right)\times 5}}{2\left(-4\right)}$$

Square 0.

$$k=\frac{0±\sqrt{-4\left(-4\right)\times 5}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$k=\frac{0±\sqrt{16\times 5}}{2\left(-4\right)}$$

Multiply 16 times 5.

$$k=\frac{0±\sqrt{80}}{2\left(-4\right)}$$

Take the square root of 80.

$$k=\frac{0±4\sqrt{5}}{2\left(-4\right)}$$

Multiply 2 times -4.

$$k=\frac{0±4\sqrt{5}}{-8}$$

Now solve the equation k=\frac{0±4\sqrt{5}}{-8} when ± is plus.

$$k=-\frac{\sqrt{5}}{2}$$

Now solve the equation k=\frac{0±4\sqrt{5}}{-8} when ± is minus.

$$k=\frac{\sqrt{5}}{2}$$

The equation is now solved.

$$k=-\frac{\sqrt{5}}{2} k=\frac{\sqrt{5}}{2}$$