Solve (2*lambda-4/4)^2+8(-lambda/4)=0

$$( 2 \cdot \frac { \lambda - 4 } { 4 } ) ^ { 2 } + 8 ( - \frac { \lambda } { 4 } ) = 0$$

Solve for λ

$$\lambda =4\sqrt{3}+8\approx 14.92820323$$

$$\lambda =8-4\sqrt{3}\approx 1.07179677$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 4.

$$4\times \left(2\times \frac{\lambda -4}{4}\right)^{2}+32\left(-\frac{\lambda }{4}\right)=0$$

Cancel out 4, the greatest common factor in 2 and 4.

$$4\times \left(\frac{\lambda -4}{2}\right)^{2}+32\left(-\frac{\lambda }{4}\right)=0$$

To raise \frac{\lambda -4}{2} to a power, raise both numerator and denominator to the power and then divide.

$$4\times \frac{\left(\lambda -4\right)^{2}}{2^{2}}+32\left(-\frac{\lambda }{4}\right)=0$$

Express 4\times \frac{\left(\lambda -4\right)^{2}}{2^{2}} as a single fraction.

$$\frac{4\left(\lambda -4\right)^{2}}{2^{2}}+32\left(-\frac{\lambda }{4}\right)=0$$

Cancel out 4, the greatest common factor in 32 and 4.

$$\frac{4\left(\lambda -4\right)^{2}}{2^{2}}-8\lambda =0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply -8\lambda times \frac{2^{2}}{2^{2}}.

$$\frac{4\left(\lambda -4\right)^{2}}{2^{2}}+\frac{-8\lambda \times 2^{2}}{2^{2}}=0$$

Since \frac{4\left(\lambda -4\right)^{2}}{2^{2}} and \frac{-8\lambda \times 2^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$\frac{4\left(\lambda -4\right)^{2}-8\lambda \times 2^{2}}{2^{2}}=0$$

Do the multiplications in 4\left(\lambda -4\right)^{2}-8\lambda \times 2^{2}.

$$\frac{4\lambda ^{2}-32\lambda +64-32\lambda }{2^{2}}=0$$

Combine like terms in 4\lambda ^{2}-32\lambda +64-32\lambda .

$$\frac{4\lambda ^{2}-64\lambda +64}{2^{2}}=0$$

Calculate 2 to the power of 2 and get 4.

$$\frac{4\lambda ^{2}-64\lambda +64}{4}=0$$

Divide each term of 4\lambda ^{2}-64\lambda +64 by 4 to get 16-16\lambda +\lambda ^{2}.

$$16-16\lambda +\lambda ^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$\lambda ^{2}-16\lambda +16=0$$

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

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

Square -16.

$$\lambda =\frac{-\left(-16\right)±\sqrt{256-4\times 16}}{2}$$

Multiply -4 times 16.

$$\lambda =\frac{-\left(-16\right)±\sqrt{256-64}}{2}$$

Add 256 to -64.

$$\lambda =\frac{-\left(-16\right)±\sqrt{192}}{2}$$

Take the square root of 192.

$$\lambda =\frac{-\left(-16\right)±8\sqrt{3}}{2}$$

The opposite of -16 is 16.

$$\lambda =\frac{16±8\sqrt{3}}{2}$$

Now solve the equation \lambda =\frac{16±8\sqrt{3}}{2} when ± is plus. Add 16 to 8\sqrt{3}.

$$\lambda =\frac{8\sqrt{3}+16}{2}$$

Divide 16+8\sqrt{3} by 2.

$$\lambda =4\sqrt{3}+8$$

Now solve the equation \lambda =\frac{16±8\sqrt{3}}{2} when ± is minus. Subtract 8\sqrt{3} from 16.

$$\lambda =\frac{16-8\sqrt{3}}{2}$$

Divide 16-8\sqrt{3} by 2.

$$\lambda =8-4\sqrt{3}$$

The equation is now solved.

$$\lambda =4\sqrt{3}+8 \lambda =8-4\sqrt{3}$$

Steps for Completing the Square

Multiply both sides of the equation by 4.

$$4\times \left(2\times \frac{\lambda -4}{4}\right)^{2}+32\left(-\frac{\lambda }{4}\right)=0$$

Cancel out 4, the greatest common factor in 2 and 4.

$$4\times \left(\frac{\lambda -4}{2}\right)^{2}+32\left(-\frac{\lambda }{4}\right)=0$$

To raise \frac{\lambda -4}{2} to a power, raise both numerator and denominator to the power and then divide.

$$4\times \frac{\left(\lambda -4\right)^{2}}{2^{2}}+32\left(-\frac{\lambda }{4}\right)=0$$

Express 4\times \frac{\left(\lambda -4\right)^{2}}{2^{2}} as a single fraction.

$$\frac{4\left(\lambda -4\right)^{2}}{2^{2}}+32\left(-\frac{\lambda }{4}\right)=0$$

Cancel out 4, the greatest common factor in 32 and 4.

$$\frac{4\left(\lambda -4\right)^{2}}{2^{2}}-8\lambda =0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply -8\lambda times \frac{2^{2}}{2^{2}}.

$$\frac{4\left(\lambda -4\right)^{2}}{2^{2}}+\frac{-8\lambda \times 2^{2}}{2^{2}}=0$$

Since \frac{4\left(\lambda -4\right)^{2}}{2^{2}} and \frac{-8\lambda \times 2^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$\frac{4\left(\lambda -4\right)^{2}-8\lambda \times 2^{2}}{2^{2}}=0$$

Do the multiplications in 4\left(\lambda -4\right)^{2}-8\lambda \times 2^{2}.

$$\frac{4\lambda ^{2}-32\lambda +64-32\lambda }{2^{2}}=0$$

Combine like terms in 4\lambda ^{2}-32\lambda +64-32\lambda .

$$\frac{4\lambda ^{2}-64\lambda +64}{2^{2}}=0$$

Calculate 2 to the power of 2 and get 4.

$$\frac{4\lambda ^{2}-64\lambda +64}{4}=0$$

Divide each term of 4\lambda ^{2}-64\lambda +64 by 4 to get 16-16\lambda +\lambda ^{2}.

$$16-16\lambda +\lambda ^{2}=0$$

Subtract 16 from both sides. Anything subtracted from zero gives its negation.

$$-16\lambda +\lambda ^{2}=-16$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$\lambda ^{2}-16\lambda =-16$$

Divide -16, the coefficient of the x term, by 2 to get -8. Then add the square of -8 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$\lambda ^{2}-16\lambda +\left(-8\right)^{2}=-16+\left(-8\right)^{2}$$

Square -8.

$$\lambda ^{2}-16\lambda +64=-16+64$$

Add -16 to 64.

$$\lambda ^{2}-16\lambda +64=48$$

Factor \lambda ^{2}-16\lambda +64. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(\lambda -8\right)^{2}=48$$

Take the square root of both sides of the equation.

$$\sqrt{\left(\lambda -8\right)^{2}}=\sqrt{48}$$

Simplify.

$$\lambda -8=4\sqrt{3} \lambda -8=-4\sqrt{3}$$

Add 8 to both sides of the equation.

$$\lambda =4\sqrt{3}+8 \lambda =8-4\sqrt{3}$$