Solve x^2+(-4/7x+4)^2=9

$$x ^ { 2 } + ( - \frac { 4 } { 7 } x + 4 ) ^ { 2 } = 9$$

Solve for x (complex solution)

$$x=\frac{112+7\sqrt{199}i}{65}\approx 1.723076923+1.519186952i$$

$$x=\frac{-7\sqrt{199}i+112}{65}\approx 1.723076923-1.519186952i$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-\frac{4}{7}x+4\right)^{2}.

$$x^{2}+\frac{16}{49}x^{2}-\frac{32}{7}x+16=9$$

Combine x^{2} and \frac{16}{49}x^{2} to get \frac{65}{49}x^{2}.

$$\frac{65}{49}x^{2}-\frac{32}{7}x+16=9$$

Subtract 9 from both sides.

$$\frac{65}{49}x^{2}-\frac{32}{7}x+16-9=0$$

Subtract 9 from 16 to get 7.

$$\frac{65}{49}x^{2}-\frac{32}{7}x+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{65}{49} for a, -\frac{32}{7} for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{32}{7}\right)±\sqrt{\left(-\frac{32}{7}\right)^{2}-4\times \frac{65}{49}\times 7}}{2\times \frac{65}{49}}$$

Square -\frac{32}{7} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{32}{7}\right)±\sqrt{\frac{1024}{49}-4\times \frac{65}{49}\times 7}}{2\times \frac{65}{49}}$$

Multiply -4 times \frac{65}{49}.

$$x=\frac{-\left(-\frac{32}{7}\right)±\sqrt{\frac{1024}{49}-\frac{260}{49}\times 7}}{2\times \frac{65}{49}}$$

Multiply -\frac{260}{49} times 7.

$$x=\frac{-\left(-\frac{32}{7}\right)±\sqrt{\frac{1024}{49}-\frac{260}{7}}}{2\times \frac{65}{49}}$$

Add \frac{1024}{49} to -\frac{260}{7} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{32}{7}\right)±\sqrt{-\frac{796}{49}}}{2\times \frac{65}{49}}$$

Take the square root of -\frac{796}{49}.

$$x=\frac{-\left(-\frac{32}{7}\right)±\frac{2\sqrt{199}i}{7}}{2\times \frac{65}{49}}$$

The opposite of -\frac{32}{7} is \frac{32}{7}.

$$x=\frac{\frac{32}{7}±\frac{2\sqrt{199}i}{7}}{2\times \frac{65}{49}}$$

Multiply 2 times \frac{65}{49}.

$$x=\frac{\frac{32}{7}±\frac{2\sqrt{199}i}{7}}{\frac{130}{49}}$$

Now solve the equation x=\frac{\frac{32}{7}±\frac{2\sqrt{199}i}{7}}{\frac{130}{49}} when ± is plus. Add \frac{32}{7} to \frac{2i\sqrt{199}}{7}.

$$x=\frac{32+2\sqrt{199}i}{\frac{130}{49}\times 7}$$

Divide \frac{32+2i\sqrt{199}}{7} by \frac{130}{49} by multiplying \frac{32+2i\sqrt{199}}{7} by the reciprocal of \frac{130}{49}.

$$x=\frac{112+7\sqrt{199}i}{65}$$

Now solve the equation x=\frac{\frac{32}{7}±\frac{2\sqrt{199}i}{7}}{\frac{130}{49}} when ± is minus. Subtract \frac{2i\sqrt{199}}{7} from \frac{32}{7}.

$$x=\frac{-2\sqrt{199}i+32}{\frac{130}{49}\times 7}$$

Divide \frac{32-2i\sqrt{199}}{7} by \frac{130}{49} by multiplying \frac{32-2i\sqrt{199}}{7} by the reciprocal of \frac{130}{49}.

$$x=\frac{-7\sqrt{199}i+112}{65}$$

The equation is now solved.

$$x=\frac{112+7\sqrt{199}i}{65} x=\frac{-7\sqrt{199}i+112}{65}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-\frac{4}{7}x+4\right)^{2}.

$$x^{2}+\frac{16}{49}x^{2}-\frac{32}{7}x+16=9$$

Combine x^{2} and \frac{16}{49}x^{2} to get \frac{65}{49}x^{2}.

$$\frac{65}{49}x^{2}-\frac{32}{7}x+16=9$$

Subtract 16 from both sides.

$$\frac{65}{49}x^{2}-\frac{32}{7}x=9-16$$

Subtract 16 from 9 to get -7.

$$\frac{65}{49}x^{2}-\frac{32}{7}x=-7$$

Divide both sides of the equation by \frac{65}{49}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{65}{49}x^{2}-\frac{32}{7}x}{\frac{65}{49}}=-\frac{7}{\frac{65}{49}}$$

Dividing by \frac{65}{49} undoes the multiplication by \frac{65}{49}.

$$x^{2}+\left(-\frac{\frac{32}{7}}{\frac{65}{49}}\right)x=-\frac{7}{\frac{65}{49}}$$

Divide -\frac{32}{7} by \frac{65}{49} by multiplying -\frac{32}{7} by the reciprocal of \frac{65}{49}.

$$x^{2}-\frac{224}{65}x=-\frac{7}{\frac{65}{49}}$$

Divide -7 by \frac{65}{49} by multiplying -7 by the reciprocal of \frac{65}{49}.

$$x^{2}-\frac{224}{65}x=-\frac{343}{65}$$

Divide -\frac{224}{65}, the coefficient of the x term, by 2 to get -\frac{112}{65}. Then add the square of -\frac{112}{65} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{224}{65}x+\left(-\frac{112}{65}\right)^{2}=-\frac{343}{65}+\left(-\frac{112}{65}\right)^{2}$$

Square -\frac{112}{65} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{224}{65}x+\frac{12544}{4225}=-\frac{343}{65}+\frac{12544}{4225}$$

Add -\frac{343}{65} to \frac{12544}{4225} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{224}{65}x+\frac{12544}{4225}=-\frac{9751}{4225}$$

Factor x^{2}-\frac{224}{65}x+\frac{12544}{4225}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{112}{65}\right)^{2}=-\frac{9751}{4225}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{112}{65}\right)^{2}}=\sqrt{-\frac{9751}{4225}}$$

Simplify.

$$x-\frac{112}{65}=\frac{7\sqrt{199}i}{65} x-\frac{112}{65}=-\frac{7\sqrt{199}i}{65}$$

Add \frac{112}{65} to both sides of the equation.

$$x=\frac{112+7\sqrt{199}i}{65} x=\frac{-7\sqrt{199}i+112}{65}$$