Solve 25/9x^2=left(x-3right)^2

$$\frac{ 25 }{ 9 } { x }^{ 2 } = { \left(x-3 \right) }^{ 2 }$$

Solve for x

$$x = -\frac{9}{2} = -4\frac{1}{2} = -4.5$$

$$x = \frac{9}{8} = 1\frac{1}{8} = 1.125$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$\frac{25}{9}x^{2}=x^{2}-6x+9$$

Subtract x^{2} from both sides.

$$\frac{25}{9}x^{2}-x^{2}=-6x+9$$

Combine \frac{25}{9}x^{2} and -x^{2} to get \frac{16}{9}x^{2}.

$$\frac{16}{9}x^{2}=-6x+9$$

Add 6x to both sides.

$$\frac{16}{9}x^{2}+6x=9$$

Subtract 9 from both sides.

$$\frac{16}{9}x^{2}+6x-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{16}{9} for a, 6 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times \frac{16}{9}\left(-9\right)}}{2\times \frac{16}{9}}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times \frac{16}{9}\left(-9\right)}}{2\times \frac{16}{9}}$$

Multiply -4 times \frac{16}{9}.

$$x=\frac{-6±\sqrt{36-\frac{64}{9}\left(-9\right)}}{2\times \frac{16}{9}}$$

Multiply -\frac{64}{9} times -9.

$$x=\frac{-6±\sqrt{36+64}}{2\times \frac{16}{9}}$$

Add 36 to 64.

$$x=\frac{-6±\sqrt{100}}{2\times \frac{16}{9}}$$

Take the square root of 100.

$$x=\frac{-6±10}{2\times \frac{16}{9}}$$

Multiply 2 times \frac{16}{9}.

$$x=\frac{-6±10}{\frac{32}{9}}$$

Now solve the equation x=\frac{-6±10}{\frac{32}{9}} when ± is plus. Add -6 to 10.

$$x=\frac{4}{\frac{32}{9}}$$

Divide 4 by \frac{32}{9} by multiplying 4 by the reciprocal of \frac{32}{9}.

$$x=\frac{9}{8}$$

Now solve the equation x=\frac{-6±10}{\frac{32}{9}} when ± is minus. Subtract 10 from -6.

$$x=-\frac{16}{\frac{32}{9}}$$

Divide -16 by \frac{32}{9} by multiplying -16 by the reciprocal of \frac{32}{9}.

$$x=-\frac{9}{2}$$

The equation is now solved.

$$x=\frac{9}{8} x=-\frac{9}{2}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$\frac{25}{9}x^{2}=x^{2}-6x+9$$

Subtract x^{2} from both sides.

$$\frac{25}{9}x^{2}-x^{2}=-6x+9$$

Combine \frac{25}{9}x^{2} and -x^{2} to get \frac{16}{9}x^{2}.

$$\frac{16}{9}x^{2}=-6x+9$$

Add 6x to both sides.

$$\frac{16}{9}x^{2}+6x=9$$

Divide both sides of the equation by \frac{16}{9}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{16}{9}x^{2}+6x}{\frac{16}{9}}=\frac{9}{\frac{16}{9}}$$

Dividing by \frac{16}{9} undoes the multiplication by \frac{16}{9}.

$$x^{2}+\frac{6}{\frac{16}{9}}x=\frac{9}{\frac{16}{9}}$$

Divide 6 by \frac{16}{9} by multiplying 6 by the reciprocal of \frac{16}{9}.

$$x^{2}+\frac{27}{8}x=\frac{9}{\frac{16}{9}}$$

Divide 9 by \frac{16}{9} by multiplying 9 by the reciprocal of \frac{16}{9}.

$$x^{2}+\frac{27}{8}x=\frac{81}{16}$$

Divide \frac{27}{8}, the coefficient of the x term, by 2 to get \frac{27}{16}. Then add the square of \frac{27}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{27}{8}x+\left(\frac{27}{16}\right)^{2}=\frac{81}{16}+\left(\frac{27}{16}\right)^{2}$$

Square \frac{27}{16} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{27}{8}x+\frac{729}{256}=\frac{81}{16}+\frac{729}{256}$$

Add \frac{81}{16} to \frac{729}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{27}{8}x+\frac{729}{256}=\frac{2025}{256}$$

Factor x^{2}+\frac{27}{8}x+\frac{729}{256}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{27}{16}\right)^{2}=\frac{2025}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{27}{16}\right)^{2}}=\sqrt{\frac{2025}{256}}$$

Simplify.

$$x+\frac{27}{16}=\frac{45}{16} x+\frac{27}{16}=-\frac{45}{16}$$

Subtract \frac{27}{16} from both sides of the equation.

$$x=\frac{9}{8} x=-\frac{9}{2}$$