Solve 1/4x^2+1+1/4x^2+{left(-2/3x^2+7/3x-1right)}^2=x^2+{left(-frac{2}{3}{x}^2+frac{7}{3}xright)}^2

$$\frac{ 1 }{ 4 } { x }^{ 2 } +1+ \frac{ 1 }{ 4 } { x }^{ 2 } + { \left(- \frac{ 2 }{ 3 } { x }^{ 2 } + \frac{ 7 }{ 3 } x-1 \right) }^{ 2 } = { x }^{ 2 } + { \left(- \frac{ 2 }{ 3 } { x }^{ 2 } + \frac{ 7 }{ 3 } x \right) }^{ 2 }$$

Solve for x

$$x = \frac{2 \sqrt{34} + 14}{5} \approx 5.132380758$$

$$x=\frac{14-2\sqrt{34}}{5}\approx 0.467619242$$

Steps Using the Quadratic Formula

Combine \frac{1}{4}x^{2} and \frac{1}{4}x^{2} to get \frac{1}{2}x^{2}.

$$\frac{1}{2}x^{2}+1+\left(-\frac{2}{3}x^{2}+\frac{7}{3}x-1\right)^{2}=x^{2}+\left(-\frac{2}{3}x^{2}+\frac{7}{3}x\right)^{2}$$

Square -\frac{2}{3}x^{2}+\frac{7}{3}x-1.

$$\frac{1}{2}x^{2}+1+\frac{4}{9}x^{4}-\frac{28}{9}x^{3}+\frac{61}{9}x^{2}-\frac{14}{3}x+1=x^{2}+\left(-\frac{2}{3}x^{2}+\frac{7}{3}x\right)^{2}$$

Combine \frac{1}{2}x^{2} and \frac{61}{9}x^{2} to get \frac{131}{18}x^{2}.

$$\frac{131}{18}x^{2}+1+\frac{4}{9}x^{4}-\frac{28}{9}x^{3}-\frac{14}{3}x+1=x^{2}+\left(-\frac{2}{3}x^{2}+\frac{7}{3}x\right)^{2}$$

Add 1 and 1 to get 2.

$$\frac{131}{18}x^{2}+2+\frac{4}{9}x^{4}-\frac{28}{9}x^{3}-\frac{14}{3}x=x^{2}+\left(-\frac{2}{3}x^{2}+\frac{7}{3}x\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-\frac{2}{3}x^{2}+\frac{7}{3}x\right)^{2}.

$$\frac{131}{18}x^{2}+2+\frac{4}{9}x^{4}-\frac{28}{9}x^{3}-\frac{14}{3}x=x^{2}+\frac{4}{9}\left(x^{2}\right)^{2}-\frac{28}{9}x^{2}x+\frac{49}{9}x^{2}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$\frac{131}{18}x^{2}+2+\frac{4}{9}x^{4}-\frac{28}{9}x^{3}-\frac{14}{3}x=x^{2}+\frac{4}{9}x^{4}-\frac{28}{9}x^{2}x+\frac{49}{9}x^{2}$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$\frac{131}{18}x^{2}+2+\frac{4}{9}x^{4}-\frac{28}{9}x^{3}-\frac{14}{3}x=x^{2}+\frac{4}{9}x^{4}-\frac{28}{9}x^{3}+\frac{49}{9}x^{2}$$

Combine x^{2} and \frac{49}{9}x^{2} to get \frac{58}{9}x^{2}.

$$\frac{131}{18}x^{2}+2+\frac{4}{9}x^{4}-\frac{28}{9}x^{3}-\frac{14}{3}x=\frac{58}{9}x^{2}+\frac{4}{9}x^{4}-\frac{28}{9}x^{3}$$

Subtract \frac{58}{9}x^{2} from both sides.

$$\frac{131}{18}x^{2}+2+\frac{4}{9}x^{4}-\frac{28}{9}x^{3}-\frac{14}{3}x-\frac{58}{9}x^{2}=\frac{4}{9}x^{4}-\frac{28}{9}x^{3}$$

Combine \frac{131}{18}x^{2} and -\frac{58}{9}x^{2} to get \frac{5}{6}x^{2}.

$$\frac{5}{6}x^{2}+2+\frac{4}{9}x^{4}-\frac{28}{9}x^{3}-\frac{14}{3}x=\frac{4}{9}x^{4}-\frac{28}{9}x^{3}$$

Subtract \frac{4}{9}x^{4} from both sides.

$$\frac{5}{6}x^{2}+2+\frac{4}{9}x^{4}-\frac{28}{9}x^{3}-\frac{14}{3}x-\frac{4}{9}x^{4}=-\frac{28}{9}x^{3}$$

Combine \frac{4}{9}x^{4} and -\frac{4}{9}x^{4} to get 0.

$$\frac{5}{6}x^{2}+2-\frac{28}{9}x^{3}-\frac{14}{3}x=-\frac{28}{9}x^{3}$$

Add \frac{28}{9}x^{3} to both sides.

$$\frac{5}{6}x^{2}+2-\frac{28}{9}x^{3}-\frac{14}{3}x+\frac{28}{9}x^{3}=0$$

Combine -\frac{28}{9}x^{3} and \frac{28}{9}x^{3} to get 0.

$$\frac{5}{6}x^{2}+2-\frac{14}{3}x=0$$

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.

$$\frac{5}{6}x^{2}-\frac{14}{3}x+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{5}{6} for a, -\frac{14}{3} for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{14}{3}\right)±\sqrt{\left(-\frac{14}{3}\right)^{2}-4\times \frac{5}{6}\times 2}}{2\times \frac{5}{6}}$$

Square -\frac{14}{3} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{14}{3}\right)±\sqrt{\frac{196}{9}-4\times \frac{5}{6}\times 2}}{2\times \frac{5}{6}}$$

Multiply -4 times \frac{5}{6}.

$$x=\frac{-\left(-\frac{14}{3}\right)±\sqrt{\frac{196}{9}-\frac{10}{3}\times 2}}{2\times \frac{5}{6}}$$

Multiply -\frac{10}{3} times 2.

$$x=\frac{-\left(-\frac{14}{3}\right)±\sqrt{\frac{196}{9}-\frac{20}{3}}}{2\times \frac{5}{6}}$$

Add \frac{196}{9} to -\frac{20}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{14}{3}\right)±\sqrt{\frac{136}{9}}}{2\times \frac{5}{6}}$$

Take the square root of \frac{136}{9}.

$$x=\frac{-\left(-\frac{14}{3}\right)±\frac{2\sqrt{34}}{3}}{2\times \frac{5}{6}}$$

The opposite of -\frac{14}{3} is \frac{14}{3}.

$$x=\frac{\frac{14}{3}±\frac{2\sqrt{34}}{3}}{2\times \frac{5}{6}}$$

Multiply 2 times \frac{5}{6}.

$$x=\frac{\frac{14}{3}±\frac{2\sqrt{34}}{3}}{\frac{5}{3}}$$

Now solve the equation x=\frac{\frac{14}{3}±\frac{2\sqrt{34}}{3}}{\frac{5}{3}} when ± is plus. Add \frac{14}{3} to \frac{2\sqrt{34}}{3}.

$$x=\frac{2\sqrt{34}+14}{\frac{5}{3}\times 3}$$

Divide \frac{14+2\sqrt{34}}{3} by \frac{5}{3} by multiplying \frac{14+2\sqrt{34}}{3} by the reciprocal of \frac{5}{3}.

$$x=\frac{2\sqrt{34}+14}{5}$$

Now solve the equation x=\frac{\frac{14}{3}±\frac{2\sqrt{34}}{3}}{\frac{5}{3}} when ± is minus. Subtract \frac{2\sqrt{34}}{3} from \frac{14}{3}.

$$x=\frac{14-2\sqrt{34}}{\frac{5}{3}\times 3}$$

Divide \frac{14-2\sqrt{34}}{3} by \frac{5}{3} by multiplying \frac{14-2\sqrt{34}}{3} by the reciprocal of \frac{5}{3}.

$$x=\frac{14-2\sqrt{34}}{5}$$

The equation is now solved.

$$x=\frac{2\sqrt{34}+14}{5} x=\frac{14-2\sqrt{34}}{5}$$

Steps for Completing the Square

Combine \frac{1}{4}x^{2} and \frac{1}{4}x^{2} to get \frac{1}{2}x^{2}.

$$\frac{1}{2}x^{2}+1+\left(-\frac{2}{3}x^{2}+\frac{7}{3}x-1\right)^{2}=x^{2}+\left(-\frac{2}{3}x^{2}+\frac{7}{3}x\right)^{2}$$

Square -\frac{2}{3}x^{2}+\frac{7}{3}x-1.

$$\frac{1}{2}x^{2}+1+\frac{4}{9}x^{4}-\frac{28}{9}x^{3}+\frac{61}{9}x^{2}-\frac{14}{3}x+1=x^{2}+\left(-\frac{2}{3}x^{2}+\frac{7}{3}x\right)^{2}$$

Combine \frac{1}{2}x^{2} and \frac{61}{9}x^{2} to get \frac{131}{18}x^{2}.

$$\frac{131}{18}x^{2}+1+\frac{4}{9}x^{4}-\frac{28}{9}x^{3}-\frac{14}{3}x+1=x^{2}+\left(-\frac{2}{3}x^{2}+\frac{7}{3}x\right)^{2}$$

Add 1 and 1 to get 2.

$$\frac{131}{18}x^{2}+2+\frac{4}{9}x^{4}-\frac{28}{9}x^{3}-\frac{14}{3}x=x^{2}+\left(-\frac{2}{3}x^{2}+\frac{7}{3}x\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-\frac{2}{3}x^{2}+\frac{7}{3}x\right)^{2}.

$$\frac{131}{18}x^{2}+2+\frac{4}{9}x^{4}-\frac{28}{9}x^{3}-\frac{14}{3}x=x^{2}+\frac{4}{9}\left(x^{2}\right)^{2}-\frac{28}{9}x^{2}x+\frac{49}{9}x^{2}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$\frac{131}{18}x^{2}+2+\frac{4}{9}x^{4}-\frac{28}{9}x^{3}-\frac{14}{3}x=x^{2}+\frac{4}{9}x^{4}-\frac{28}{9}x^{2}x+\frac{49}{9}x^{2}$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$\frac{131}{18}x^{2}+2+\frac{4}{9}x^{4}-\frac{28}{9}x^{3}-\frac{14}{3}x=x^{2}+\frac{4}{9}x^{4}-\frac{28}{9}x^{3}+\frac{49}{9}x^{2}$$

Combine x^{2} and \frac{49}{9}x^{2} to get \frac{58}{9}x^{2}.

$$\frac{131}{18}x^{2}+2+\frac{4}{9}x^{4}-\frac{28}{9}x^{3}-\frac{14}{3}x=\frac{58}{9}x^{2}+\frac{4}{9}x^{4}-\frac{28}{9}x^{3}$$

Subtract \frac{58}{9}x^{2} from both sides.

$$\frac{131}{18}x^{2}+2+\frac{4}{9}x^{4}-\frac{28}{9}x^{3}-\frac{14}{3}x-\frac{58}{9}x^{2}=\frac{4}{9}x^{4}-\frac{28}{9}x^{3}$$

Combine \frac{131}{18}x^{2} and -\frac{58}{9}x^{2} to get \frac{5}{6}x^{2}.

$$\frac{5}{6}x^{2}+2+\frac{4}{9}x^{4}-\frac{28}{9}x^{3}-\frac{14}{3}x=\frac{4}{9}x^{4}-\frac{28}{9}x^{3}$$

Subtract \frac{4}{9}x^{4} from both sides.

$$\frac{5}{6}x^{2}+2+\frac{4}{9}x^{4}-\frac{28}{9}x^{3}-\frac{14}{3}x-\frac{4}{9}x^{4}=-\frac{28}{9}x^{3}$$

Combine \frac{4}{9}x^{4} and -\frac{4}{9}x^{4} to get 0.

$$\frac{5}{6}x^{2}+2-\frac{28}{9}x^{3}-\frac{14}{3}x=-\frac{28}{9}x^{3}$$

Add \frac{28}{9}x^{3} to both sides.

$$\frac{5}{6}x^{2}+2-\frac{28}{9}x^{3}-\frac{14}{3}x+\frac{28}{9}x^{3}=0$$

Combine -\frac{28}{9}x^{3} and \frac{28}{9}x^{3} to get 0.

$$\frac{5}{6}x^{2}+2-\frac{14}{3}x=0$$

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

$$\frac{5}{6}x^{2}-\frac{14}{3}x=-2$$

Divide both sides of the equation by \frac{5}{6}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{5}{6}x^{2}-\frac{14}{3}x}{\frac{5}{6}}=-\frac{2}{\frac{5}{6}}$$

Dividing by \frac{5}{6} undoes the multiplication by \frac{5}{6}.

$$x^{2}+\left(-\frac{\frac{14}{3}}{\frac{5}{6}}\right)x=-\frac{2}{\frac{5}{6}}$$

Divide -\frac{14}{3} by \frac{5}{6} by multiplying -\frac{14}{3} by the reciprocal of \frac{5}{6}.

$$x^{2}-\frac{28}{5}x=-\frac{2}{\frac{5}{6}}$$

Divide -2 by \frac{5}{6} by multiplying -2 by the reciprocal of \frac{5}{6}.

$$x^{2}-\frac{28}{5}x=-\frac{12}{5}$$

Divide -\frac{28}{5}, the coefficient of the x term, by 2 to get -\frac{14}{5}. Then add the square of -\frac{14}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{28}{5}x+\left(-\frac{14}{5}\right)^{2}=-\frac{12}{5}+\left(-\frac{14}{5}\right)^{2}$$

Square -\frac{14}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{28}{5}x+\frac{196}{25}=-\frac{12}{5}+\frac{196}{25}$$

Add -\frac{12}{5} to \frac{196}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{28}{5}x+\frac{196}{25}=\frac{136}{25}$$

Factor x^{2}-\frac{28}{5}x+\frac{196}{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}.

$$\left(x-\frac{14}{5}\right)^{2}=\frac{136}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{14}{5}\right)^{2}}=\sqrt{\frac{136}{25}}$$

Simplify.

$$x-\frac{14}{5}=\frac{2\sqrt{34}}{5} x-\frac{14}{5}=-\frac{2\sqrt{34}}{5}$$

Add \frac{14}{5} to both sides of the equation.

$$x=\frac{2\sqrt{34}+14}{5} x=\frac{14-2\sqrt{34}}{5}$$