Solve [4/3(x-2)]^2=2x

$$[ \frac { 4 } { 3 } ( x - 2 ) ] ^ { 2 } = 2 x$$

Solve for x

$$x = \frac{3 \sqrt{73} + 41}{16} \approx 4.164500702$$

$$x=\frac{41-3\sqrt{73}}{16}\approx 0.960499298$$

Steps Using the Quadratic Formula

Use the distributive property to multiply \frac{4}{3} by x-2.

$$\left(\frac{4}{3}x-\frac{8}{3}\right)^{2}=2x$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{4}{3}x-\frac{8}{3}\right)^{2}.

$$\frac{16}{9}x^{2}-\frac{64}{9}x+\frac{64}{9}=2x$$

Subtract 2x from both sides.

$$\frac{16}{9}x^{2}-\frac{64}{9}x+\frac{64}{9}-2x=0$$

Combine -\frac{64}{9}x and -2x to get -\frac{82}{9}x.

$$\frac{16}{9}x^{2}-\frac{82}{9}x+\frac{64}{9}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{16}{9} for a, -\frac{82}{9} for b, and \frac{64}{9} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{82}{9}\right)±\sqrt{\left(-\frac{82}{9}\right)^{2}-4\times \frac{16}{9}\times \frac{64}{9}}}{2\times \frac{16}{9}}$$

Square -\frac{82}{9} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{82}{9}\right)±\sqrt{\frac{6724}{81}-4\times \frac{16}{9}\times \frac{64}{9}}}{2\times \frac{16}{9}}$$

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

$$x=\frac{-\left(-\frac{82}{9}\right)±\sqrt{\frac{6724}{81}-\frac{64}{9}\times \frac{64}{9}}}{2\times \frac{16}{9}}$$

Multiply -\frac{64}{9} times \frac{64}{9} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{82}{9}\right)±\sqrt{\frac{6724-4096}{81}}}{2\times \frac{16}{9}}$$

Add \frac{6724}{81} to -\frac{4096}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{82}{9}\right)±\sqrt{\frac{292}{9}}}{2\times \frac{16}{9}}$$

Take the square root of \frac{292}{9}.

$$x=\frac{-\left(-\frac{82}{9}\right)±\frac{2\sqrt{73}}{3}}{2\times \frac{16}{9}}$$

The opposite of -\frac{82}{9} is \frac{82}{9}.

$$x=\frac{\frac{82}{9}±\frac{2\sqrt{73}}{3}}{2\times \frac{16}{9}}$$

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

$$x=\frac{\frac{82}{9}±\frac{2\sqrt{73}}{3}}{\frac{32}{9}}$$

Now solve the equation x=\frac{\frac{82}{9}±\frac{2\sqrt{73}}{3}}{\frac{32}{9}} when ± is plus. Add \frac{82}{9} to \frac{2\sqrt{73}}{3}.

$$x=\frac{\frac{2\sqrt{73}}{3}+\frac{82}{9}}{\frac{32}{9}}$$

Divide \frac{82}{9}+\frac{2\sqrt{73}}{3} by \frac{32}{9} by multiplying \frac{82}{9}+\frac{2\sqrt{73}}{3} by the reciprocal of \frac{32}{9}.

$$x=\frac{3\sqrt{73}+41}{16}$$

Now solve the equation x=\frac{\frac{82}{9}±\frac{2\sqrt{73}}{3}}{\frac{32}{9}} when ± is minus. Subtract \frac{2\sqrt{73}}{3} from \frac{82}{9}.

$$x=\frac{-\frac{2\sqrt{73}}{3}+\frac{82}{9}}{\frac{32}{9}}$$

Divide \frac{82}{9}-\frac{2\sqrt{73}}{3} by \frac{32}{9} by multiplying \frac{82}{9}-\frac{2\sqrt{73}}{3} by the reciprocal of \frac{32}{9}.

$$x=\frac{41-3\sqrt{73}}{16}$$

The equation is now solved.

$$x=\frac{3\sqrt{73}+41}{16} x=\frac{41-3\sqrt{73}}{16}$$

Steps for Completing the Square

Use the distributive property to multiply \frac{4}{3} by x-2.

$$\left(\frac{4}{3}x-\frac{8}{3}\right)^{2}=2x$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{4}{3}x-\frac{8}{3}\right)^{2}.

$$\frac{16}{9}x^{2}-\frac{64}{9}x+\frac{64}{9}=2x$$

Subtract 2x from both sides.

$$\frac{16}{9}x^{2}-\frac{64}{9}x+\frac{64}{9}-2x=0$$

Combine -\frac{64}{9}x and -2x to get -\frac{82}{9}x.

$$\frac{16}{9}x^{2}-\frac{82}{9}x+\frac{64}{9}=0$$

Subtract \frac{64}{9} from both sides. Anything subtracted from zero gives its negation.

$$\frac{16}{9}x^{2}-\frac{82}{9}x=-\frac{64}{9}$$

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

$$\frac{\frac{16}{9}x^{2}-\frac{82}{9}x}{\frac{16}{9}}=-\frac{\frac{64}{9}}{\frac{16}{9}}$$

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

$$x^{2}+\left(-\frac{\frac{82}{9}}{\frac{16}{9}}\right)x=-\frac{\frac{64}{9}}{\frac{16}{9}}$$

Divide -\frac{82}{9} by \frac{16}{9} by multiplying -\frac{82}{9} by the reciprocal of \frac{16}{9}.

$$x^{2}-\frac{41}{8}x=-\frac{\frac{64}{9}}{\frac{16}{9}}$$

Divide -\frac{64}{9} by \frac{16}{9} by multiplying -\frac{64}{9} by the reciprocal of \frac{16}{9}.

$$x^{2}-\frac{41}{8}x=-4$$

Divide -\frac{41}{8}, the coefficient of the x term, by 2 to get -\frac{41}{16}. Then add the square of -\frac{41}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{41}{8}x+\left(-\frac{41}{16}\right)^{2}=-4+\left(-\frac{41}{16}\right)^{2}$$

Square -\frac{41}{16} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{41}{8}x+\frac{1681}{256}=-4+\frac{1681}{256}$$

Add -4 to \frac{1681}{256}.

$$x^{2}-\frac{41}{8}x+\frac{1681}{256}=\frac{657}{256}$$

Factor x^{2}-\frac{41}{8}x+\frac{1681}{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}.

$$\left(x-\frac{41}{16}\right)^{2}=\frac{657}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{41}{16}\right)^{2}}=\sqrt{\frac{657}{256}}$$

Simplify.

$$x-\frac{41}{16}=\frac{3\sqrt{73}}{16} x-\frac{41}{16}=-\frac{3\sqrt{73}}{16}$$

Add \frac{41}{16} to both sides of the equation.

$$x=\frac{3\sqrt{73}+41}{16} x=\frac{41-3\sqrt{73}}{16}$$