Solve (7/3-lambda)*(1-lambda)-(frac{2sqrt{3}}{3})^2=0

$$( \frac { 7 } { 3 } - \lambda ) \times ( 1 - \lambda ) - ( \frac { 2 \sqrt { 3 } } { 3 } ) ^ { 2 } = 0$$

Solve for λ

$$\lambda =3$$

$$\lambda =\frac{1}{3}\approx 0.333333333$$

Steps Using the Quadratic Formula

Use the distributive property to multiply \frac{7}{3}-\lambda by 1-\lambda and combine like terms.

$$\frac{7}{3}-\frac{10}{3}\lambda +\lambda ^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}=0$$

To raise \frac{2\sqrt{3}}{3} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{7}{3}-\frac{10}{3}\lambda +\lambda ^{2}-\frac{\left(2\sqrt{3}\right)^{2}}{3^{2}}=0$$

Expand \left(2\sqrt{3}\right)^{2}.

$$\frac{7}{3}-\frac{10}{3}\lambda +\lambda ^{2}-\frac{2^{2}\left(\sqrt{3}\right)^{2}}{3^{2}}=0$$

Calculate 2 to the power of 2 and get 4.

$$\frac{7}{3}-\frac{10}{3}\lambda +\lambda ^{2}-\frac{4\left(\sqrt{3}\right)^{2}}{3^{2}}=0$$

The square of \sqrt{3} is 3.

$$\frac{7}{3}-\frac{10}{3}\lambda +\lambda ^{2}-\frac{4\times 3}{3^{2}}=0$$

Multiply 4 and 3 to get 12.

$$\frac{7}{3}-\frac{10}{3}\lambda +\lambda ^{2}-\frac{12}{3^{2}}=0$$

Calculate 3 to the power of 2 and get 9.

$$\frac{7}{3}-\frac{10}{3}\lambda +\lambda ^{2}-\frac{12}{9}=0$$

Reduce the fraction \frac{12}{9} to lowest terms by extracting and canceling out 3.

$$\frac{7}{3}-\frac{10}{3}\lambda +\lambda ^{2}-\frac{4}{3}=0$$

Subtract \frac{4}{3} from \frac{7}{3} to get 1.

$$1-\frac{10}{3}\lambda +\lambda ^{2}=0$$

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.

$$\lambda ^{2}-\frac{10}{3}\lambda +1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -\frac{10}{3} for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$\lambda =\frac{-\left(-\frac{10}{3}\right)±\sqrt{\left(-\frac{10}{3}\right)^{2}-4}}{2}$$

Square -\frac{10}{3} by squaring both the numerator and the denominator of the fraction.

$$\lambda =\frac{-\left(-\frac{10}{3}\right)±\sqrt{\frac{100}{9}-4}}{2}$$

Add \frac{100}{9} to -4.

$$\lambda =\frac{-\left(-\frac{10}{3}\right)±\sqrt{\frac{64}{9}}}{2}$$

Take the square root of \frac{64}{9}.

$$\lambda =\frac{-\left(-\frac{10}{3}\right)±\frac{8}{3}}{2}$$

The opposite of -\frac{10}{3} is \frac{10}{3}.

$$\lambda =\frac{\frac{10}{3}±\frac{8}{3}}{2}$$

Now solve the equation \lambda =\frac{\frac{10}{3}±\frac{8}{3}}{2} when ± is plus. Add \frac{10}{3} to \frac{8}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$\lambda =\frac{6}{2}$$

Divide 6 by 2.

$$\lambda =3$$

Now solve the equation \lambda =\frac{\frac{10}{3}±\frac{8}{3}}{2} when ± is minus. Subtract \frac{8}{3} from \frac{10}{3} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$\lambda =\frac{\frac{2}{3}}{2}$$

Divide \frac{2}{3} by 2.

$$\lambda =\frac{1}{3}$$

The equation is now solved.

$$\lambda =3 \lambda =\frac{1}{3}$$

Steps for Completing the Square

Use the distributive property to multiply \frac{7}{3}-\lambda by 1-\lambda and combine like terms.

$$\frac{7}{3}-\frac{10}{3}\lambda +\lambda ^{2}-\left(\frac{2\sqrt{3}}{3}\right)^{2}=0$$

To raise \frac{2\sqrt{3}}{3} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{7}{3}-\frac{10}{3}\lambda +\lambda ^{2}-\frac{\left(2\sqrt{3}\right)^{2}}{3^{2}}=0$$

Expand \left(2\sqrt{3}\right)^{2}.

$$\frac{7}{3}-\frac{10}{3}\lambda +\lambda ^{2}-\frac{2^{2}\left(\sqrt{3}\right)^{2}}{3^{2}}=0$$

Calculate 2 to the power of 2 and get 4.

$$\frac{7}{3}-\frac{10}{3}\lambda +\lambda ^{2}-\frac{4\left(\sqrt{3}\right)^{2}}{3^{2}}=0$$

The square of \sqrt{3} is 3.

$$\frac{7}{3}-\frac{10}{3}\lambda +\lambda ^{2}-\frac{4\times 3}{3^{2}}=0$$

Multiply 4 and 3 to get 12.

$$\frac{7}{3}-\frac{10}{3}\lambda +\lambda ^{2}-\frac{12}{3^{2}}=0$$

Calculate 3 to the power of 2 and get 9.

$$\frac{7}{3}-\frac{10}{3}\lambda +\lambda ^{2}-\frac{12}{9}=0$$

Reduce the fraction \frac{12}{9} to lowest terms by extracting and canceling out 3.

$$\frac{7}{3}-\frac{10}{3}\lambda +\lambda ^{2}-\frac{4}{3}=0$$

Subtract \frac{4}{3} from \frac{7}{3} to get 1.

$$1-\frac{10}{3}\lambda +\lambda ^{2}=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$-\frac{10}{3}\lambda +\lambda ^{2}=-1$$

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.

$$\lambda ^{2}-\frac{10}{3}\lambda =-1$$

Divide -\frac{10}{3}, the coefficient of the x term, by 2 to get -\frac{5}{3}. Then add the square of -\frac{5}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$\lambda ^{2}-\frac{10}{3}\lambda +\left(-\frac{5}{3}\right)^{2}=-1+\left(-\frac{5}{3}\right)^{2}$$

Square -\frac{5}{3} by squaring both the numerator and the denominator of the fraction.

$$\lambda ^{2}-\frac{10}{3}\lambda +\frac{25}{9}=-1+\frac{25}{9}$$

Add -1 to \frac{25}{9}.

$$\lambda ^{2}-\frac{10}{3}\lambda +\frac{25}{9}=\frac{16}{9}$$

Factor \lambda ^{2}-\frac{10}{3}\lambda +\frac{25}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(\lambda -\frac{5}{3}\right)^{2}=\frac{16}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(\lambda -\frac{5}{3}\right)^{2}}=\sqrt{\frac{16}{9}}$$

Simplify.

$$\lambda -\frac{5}{3}=\frac{4}{3} \lambda -\frac{5}{3}=-\frac{4}{3}$$

Add \frac{5}{3} to both sides of the equation.

$$\lambda =3 \lambda =\frac{1}{3}$$