Solve (x-1)^2+(4/3x-3)^2=1

$$( x - 1 ) ^ { 2 } + ( \frac { 4 } { 3 } x - 3 ) ^ { 2 } = 1$$

Solve for x

$$x = \frac{9}{5} = 1\frac{4}{5} = 1.8$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$x^{2}-2x+1+\left(\frac{4}{3}x-3\right)^{2}=1$$

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

$$x^{2}-2x+1+\frac{16}{9}x^{2}-8x+9=1$$

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

$$\frac{25}{9}x^{2}-2x+1-8x+9=1$$

Combine -2x and -8x to get -10x.

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

Add 1 and 9 to get 10.

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

Subtract 1 from both sides.

$$\frac{25}{9}x^{2}-10x+10-1=0$$

Subtract 1 from 10 to get 9.

$$\frac{25}{9}x^{2}-10x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{25}{9} for a, -10 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times \frac{25}{9}\times 9}}{2\times \frac{25}{9}}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times \frac{25}{9}\times 9}}{2\times \frac{25}{9}}$$

Multiply -4 times \frac{25}{9}.

$$x=\frac{-\left(-10\right)±\sqrt{100-\frac{100}{9}\times 9}}{2\times \frac{25}{9}}$$

Multiply -\frac{100}{9} times 9.

$$x=\frac{-\left(-10\right)±\sqrt{100-100}}{2\times \frac{25}{9}}$$

Add 100 to -100.

$$x=\frac{-\left(-10\right)±\sqrt{0}}{2\times \frac{25}{9}}$$

Take the square root of 0.

$$x=-\frac{-10}{2\times \frac{25}{9}}$$

The opposite of -10 is 10.

$$x=\frac{10}{2\times \frac{25}{9}}$$

Multiply 2 times \frac{25}{9}.

$$x=\frac{10}{\frac{50}{9}}$$

Divide 10 by \frac{50}{9} by multiplying 10 by the reciprocal of \frac{50}{9}.

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

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$x^{2}-2x+1+\left(\frac{4}{3}x-3\right)^{2}=1$$

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

$$x^{2}-2x+1+\frac{16}{9}x^{2}-8x+9=1$$

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

$$\frac{25}{9}x^{2}-2x+1-8x+9=1$$

Combine -2x and -8x to get -10x.

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

Add 1 and 9 to get 10.

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

Subtract 10 from both sides.

$$\frac{25}{9}x^{2}-10x=1-10$$

Subtract 10 from 1 to get -9.

$$\frac{25}{9}x^{2}-10x=-9$$

Divide both sides of the equation by \frac{25}{9}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{25}{9}x^{2}-10x}{\frac{25}{9}}=-\frac{9}{\frac{25}{9}}$$

Dividing by \frac{25}{9} undoes the multiplication by \frac{25}{9}.

$$x^{2}+\left(-\frac{10}{\frac{25}{9}}\right)x=-\frac{9}{\frac{25}{9}}$$

Divide -10 by \frac{25}{9} by multiplying -10 by the reciprocal of \frac{25}{9}.

$$x^{2}-\frac{18}{5}x=-\frac{9}{\frac{25}{9}}$$

Divide -9 by \frac{25}{9} by multiplying -9 by the reciprocal of \frac{25}{9}.

$$x^{2}-\frac{18}{5}x=-\frac{81}{25}$$

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

$$x^{2}-\frac{18}{5}x+\left(-\frac{9}{5}\right)^{2}=-\frac{81}{25}+\left(-\frac{9}{5}\right)^{2}$$

Square -\frac{9}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{18}{5}x+\frac{81}{25}=\frac{-81+81}{25}$$

Add -\frac{81}{25} to \frac{81}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{18}{5}x+\frac{81}{25}=0$$

Factor x^{2}-\frac{18}{5}x+\frac{81}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{9}{5}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{5}\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-\frac{9}{5}=0 x-\frac{9}{5}=0$$

Add \frac{9}{5} to both sides of the equation.

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

The equation is now solved. Solutions are the same.

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