Solve 52/9x^2-12x+9=x^2-6x+9

$$\frac{ 52 }{ 9 } { x }^{ 2 } -12x+9= { x }^{ 2 } -6x+9$$

Solve for x

$$x = \frac{54}{43} = 1\frac{11}{43} \approx 1.255813953$$

$$x=0$$

Steps Using Factoring

Subtract x^{2} from both sides.

$$\frac{52}{9}x^{2}-12x+9-x^{2}=-6x+9$$

Combine \frac{52}{9}x^{2} and -x^{2} to get \frac{43}{9}x^{2}.

$$\frac{43}{9}x^{2}-12x+9=-6x+9$$

Add 6x to both sides.

$$\frac{43}{9}x^{2}-12x+9+6x=9$$

Combine -12x and 6x to get -6x.

$$\frac{43}{9}x^{2}-6x+9=9$$

Subtract 9 from both sides.

$$\frac{43}{9}x^{2}-6x+9-9=0$$

Subtract 9 from 9 to get 0.

$$\frac{43}{9}x^{2}-6x=0$$

Factor out x.

$$x\left(\frac{43}{9}x-6\right)=0$$

To find equation solutions, solve x=0 and \frac{43x}{9}-6=0.

$$x=0 x=\frac{54}{43}$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

$$\frac{52}{9}x^{2}-12x+9-x^{2}=-6x+9$$

Combine \frac{52}{9}x^{2} and -x^{2} to get \frac{43}{9}x^{2}.

$$\frac{43}{9}x^{2}-12x+9=-6x+9$$

Add 6x to both sides.

$$\frac{43}{9}x^{2}-12x+9+6x=9$$

Combine -12x and 6x to get -6x.

$$\frac{43}{9}x^{2}-6x+9=9$$

Subtract 9 from both sides.

$$\frac{43}{9}x^{2}-6x+9-9=0$$

Subtract 9 from 9 to get 0.

$$\frac{43}{9}x^{2}-6x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{43}{9} for a, -6 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}}}{2\times \frac{43}{9}}$$

Take the square root of \left(-6\right)^{2}.

$$x=\frac{-\left(-6\right)±6}{2\times \frac{43}{9}}$$

The opposite of -6 is 6.

$$x=\frac{6±6}{2\times \frac{43}{9}}$$

Multiply 2 times \frac{43}{9}.

$$x=\frac{6±6}{\frac{86}{9}}$$

Now solve the equation x=\frac{6±6}{\frac{86}{9}} when ± is plus. Add 6 to 6.

$$x=\frac{12}{\frac{86}{9}}$$

Divide 12 by \frac{86}{9} by multiplying 12 by the reciprocal of \frac{86}{9}.

$$x=\frac{54}{43}$$

Now solve the equation x=\frac{6±6}{\frac{86}{9}} when ± is minus. Subtract 6 from 6.

$$x=\frac{0}{\frac{86}{9}}$$

Divide 0 by \frac{86}{9} by multiplying 0 by the reciprocal of \frac{86}{9}.

$$x=0$$

The equation is now solved.

$$x=\frac{54}{43} x=0$$

Steps for Completing the Square

Subtract x^{2} from both sides.

$$\frac{52}{9}x^{2}-12x+9-x^{2}=-6x+9$$

Combine \frac{52}{9}x^{2} and -x^{2} to get \frac{43}{9}x^{2}.

$$\frac{43}{9}x^{2}-12x+9=-6x+9$$

Add 6x to both sides.

$$\frac{43}{9}x^{2}-12x+9+6x=9$$

Combine -12x and 6x to get -6x.

$$\frac{43}{9}x^{2}-6x+9=9$$

Subtract 9 from both sides.

$$\frac{43}{9}x^{2}-6x=9-9$$

Subtract 9 from 9 to get 0.

$$\frac{43}{9}x^{2}-6x=0$$

Divide both sides of the equation by \frac{43}{9}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{43}{9}x^{2}-6x}{\frac{43}{9}}=\frac{0}{\frac{43}{9}}$$

Dividing by \frac{43}{9} undoes the multiplication by \frac{43}{9}.

$$x^{2}+\left(-\frac{6}{\frac{43}{9}}\right)x=\frac{0}{\frac{43}{9}}$$

Divide -6 by \frac{43}{9} by multiplying -6 by the reciprocal of \frac{43}{9}.

$$x^{2}-\frac{54}{43}x=\frac{0}{\frac{43}{9}}$$

Divide 0 by \frac{43}{9} by multiplying 0 by the reciprocal of \frac{43}{9}.

$$x^{2}-\frac{54}{43}x=0$$

Divide -\frac{54}{43}, the coefficient of the x term, by 2 to get -\frac{27}{43}. Then add the square of -\frac{27}{43} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{54}{43}x+\left(-\frac{27}{43}\right)^{2}=\left(-\frac{27}{43}\right)^{2}$$

Square -\frac{27}{43} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{54}{43}x+\frac{729}{1849}=\frac{729}{1849}$$

Factor x^{2}-\frac{54}{43}x+\frac{729}{1849}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{27}{43}\right)^{2}=\frac{729}{1849}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{27}{43}\right)^{2}}=\sqrt{\frac{729}{1849}}$$

Simplify.

$$x-\frac{27}{43}=\frac{27}{43} x-\frac{27}{43}=-\frac{27}{43}$$

Add \frac{27}{43} to both sides of the equation.

$$x=\frac{54}{43} x=0$$