Solve 10beta*33/(beta*33-045)^2=9*33*2/(33-045)^2

$$\frac { 10 \beta \times 33 } { ( \beta \times 33 - 045 ) ^ { 2 } } = \frac { 9 \times 33 \times 2 } { ( 33 - 045 ) ^ { 2 } }$$

Solve for β

$$\beta =\frac{5}{9}\approx 0.555555556$$

Steps Using Factoring

Variable \beta cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 1089\beta ^{2}.

$$10\beta \times 33=\beta ^{2}\times 9\times 33\times 2$$

Multiply 10 and 33 to get 330.

$$330\beta =\beta ^{2}\times 9\times 33\times 2$$

Multiply 9 and 33 to get 297.

$$330\beta =\beta ^{2}\times 297\times 2$$

Multiply 297 and 2 to get 594.

$$330\beta =\beta ^{2}\times 594$$

Subtract \beta ^{2}\times 594 from both sides.

$$330\beta -\beta ^{2}\times 594=0$$

Multiply -1 and 594 to get -594.

$$330\beta -594\beta ^{2}=0$$

Factor out \beta .

$$\beta \left(330-594\beta \right)=0$$

To find equation solutions, solve \beta =0 and 330-594\beta =0.

$$\beta =0 \beta =\frac{5}{9}$$

Variable \beta cannot be equal to 0.

$$\beta =\frac{5}{9}$$

Steps Using the Quadratic Formula

Variable \beta cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 1089\beta ^{2}.

$$10\beta \times 33=\beta ^{2}\times 9\times 33\times 2$$

Multiply 10 and 33 to get 330.

$$330\beta =\beta ^{2}\times 9\times 33\times 2$$

Multiply 9 and 33 to get 297.

$$330\beta =\beta ^{2}\times 297\times 2$$

Multiply 297 and 2 to get 594.

$$330\beta =\beta ^{2}\times 594$$

Subtract \beta ^{2}\times 594 from both sides.

$$330\beta -\beta ^{2}\times 594=0$$

Multiply -1 and 594 to get -594.

$$330\beta -594\beta ^{2}=0$$

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.

$$-594\beta ^{2}+330\beta =0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -594 for a, 330 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$\beta =\frac{-330±\sqrt{330^{2}}}{2\left(-594\right)}$$

Take the square root of 330^{2}.

$$\beta =\frac{-330±330}{2\left(-594\right)}$$

Multiply 2 times -594.

$$\beta =\frac{-330±330}{-1188}$$

Now solve the equation \beta =\frac{-330±330}{-1188} when ± is plus. Add -330 to 330.

$$\beta =\frac{0}{-1188}$$

Divide 0 by -1188.

$$\beta =0$$

Now solve the equation \beta =\frac{-330±330}{-1188} when ± is minus. Subtract 330 from -330.

$$\beta =-\frac{660}{-1188}$$

Reduce the fraction \frac{-660}{-1188} to lowest terms by extracting and canceling out 132.

$$\beta =\frac{5}{9}$$

The equation is now solved.

$$\beta =0 \beta =\frac{5}{9}$$

Variable \beta cannot be equal to 0.

$$\beta =\frac{5}{9}$$

Steps for Completing the Square

Variable \beta cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 1089\beta ^{2}.

$$10\beta \times 33=\beta ^{2}\times 9\times 33\times 2$$

Multiply 10 and 33 to get 330.

$$330\beta =\beta ^{2}\times 9\times 33\times 2$$

Multiply 9 and 33 to get 297.

$$330\beta =\beta ^{2}\times 297\times 2$$

Multiply 297 and 2 to get 594.

$$330\beta =\beta ^{2}\times 594$$

Subtract \beta ^{2}\times 594 from both sides.

$$330\beta -\beta ^{2}\times 594=0$$

Multiply -1 and 594 to get -594.

$$330\beta -594\beta ^{2}=0$$

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.

$$-594\beta ^{2}+330\beta =0$$

Divide both sides by -594.

$$\frac{-594\beta ^{2}+330\beta }{-594}=\frac{0}{-594}$$

Dividing by -594 undoes the multiplication by -594.

$$\beta ^{2}+\frac{330}{-594}\beta =\frac{0}{-594}$$

Reduce the fraction \frac{330}{-594} to lowest terms by extracting and canceling out 66.

$$\beta ^{2}-\frac{5}{9}\beta =\frac{0}{-594}$$

Divide 0 by -594.

$$\beta ^{2}-\frac{5}{9}\beta =0$$

Divide -\frac{5}{9}, the coefficient of the x term, by 2 to get -\frac{5}{18}. Then add the square of -\frac{5}{18} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$\beta ^{2}-\frac{5}{9}\beta +\left(-\frac{5}{18}\right)^{2}=\left(-\frac{5}{18}\right)^{2}$$

Square -\frac{5}{18} by squaring both the numerator and the denominator of the fraction.

$$\beta ^{2}-\frac{5}{9}\beta +\frac{25}{324}=\frac{25}{324}$$

Factor \beta ^{2}-\frac{5}{9}\beta +\frac{25}{324}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(\beta -\frac{5}{18}\right)^{2}=\frac{25}{324}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(\beta -\frac{5}{18}\right)^{2}}=\sqrt{\frac{25}{324}}$$

Simplify.

$$\beta -\frac{5}{18}=\frac{5}{18} \beta -\frac{5}{18}=-\frac{5}{18}$$

Add \frac{5}{18} to both sides of the equation.

$$\beta =\frac{5}{9} \beta =0$$

Variable \beta cannot be equal to 0.

$$\beta =\frac{5}{9}$$