Solve (14-7x/3)^2+(x-3)^2=(6-x)^2

$$( \frac { 14 - 7 x } { 3 } ) ^ { 2 } + ( x - 3 ) ^ { 2 } = ( 6 - x ) ^ { 2 }$$

Solve for x

$$x = \frac{12 \sqrt{51} + 71}{49} \approx 3.19790084$$

$$x=\frac{71-12\sqrt{51}}{49}\approx -0.299941656$$

Steps Using the Quadratic Formula

To raise \frac{14-7x}{3} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(14-7x\right)^{2}}{3^{2}}+\left(x-3\right)^{2}=\left(6-x\right)^{2}$$

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

$$\frac{\left(14-7x\right)^{2}}{3^{2}}+x^{2}-6x+9=\left(6-x\right)^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{2}-6x+9 times \frac{3^{2}}{3^{2}}.

$$\frac{\left(14-7x\right)^{2}}{3^{2}}+\frac{\left(x^{2}-6x+9\right)\times 3^{2}}{3^{2}}=\left(6-x\right)^{2}$$

Since \frac{\left(14-7x\right)^{2}}{3^{2}} and \frac{\left(x^{2}-6x+9\right)\times 3^{2}}{3^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(14-7x\right)^{2}+\left(x^{2}-6x+9\right)\times 3^{2}}{3^{2}}=\left(6-x\right)^{2}$$

Do the multiplications in \left(14-7x\right)^{2}+\left(x^{2}-6x+9\right)\times 3^{2}.

$$\frac{196-196x+49x^{2}+9x^{2}-54x+81}{3^{2}}=\left(6-x\right)^{2}$$

Combine like terms in 196-196x+49x^{2}+9x^{2}-54x+81.

$$\frac{277-250x+58x^{2}}{3^{2}}=\left(6-x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6-x\right)^{2}.

$$\frac{277-250x+58x^{2}}{3^{2}}=36-12x+x^{2}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{277-250x+58x^{2}}{9}=36-12x+x^{2}$$

Divide each term of 277-250x+58x^{2} by 9 to get \frac{277}{9}-\frac{250}{9}x+\frac{58}{9}x^{2}.

$$\frac{277}{9}-\frac{250}{9}x+\frac{58}{9}x^{2}=36-12x+x^{2}$$

Subtract 36 from both sides.

$$\frac{277}{9}-\frac{250}{9}x+\frac{58}{9}x^{2}-36=-12x+x^{2}$$

Subtract 36 from \frac{277}{9} to get -\frac{47}{9}.

$$-\frac{47}{9}-\frac{250}{9}x+\frac{58}{9}x^{2}=-12x+x^{2}$$

Add 12x to both sides.

$$-\frac{47}{9}-\frac{250}{9}x+\frac{58}{9}x^{2}+12x=x^{2}$$

Combine -\frac{250}{9}x and 12x to get -\frac{142}{9}x.

$$-\frac{47}{9}-\frac{142}{9}x+\frac{58}{9}x^{2}=x^{2}$$

Subtract x^{2} from both sides.

$$-\frac{47}{9}-\frac{142}{9}x+\frac{58}{9}x^{2}-x^{2}=0$$

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

$$-\frac{47}{9}-\frac{142}{9}x+\frac{49}{9}x^{2}=0$$

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.

$$\frac{49}{9}x^{2}-\frac{142}{9}x-\frac{47}{9}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{49}{9} for a, -\frac{142}{9} for b, and -\frac{47}{9} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{142}{9}\right)±\sqrt{\left(-\frac{142}{9}\right)^{2}-4\times \frac{49}{9}\left(-\frac{47}{9}\right)}}{2\times \frac{49}{9}}$$

Square -\frac{142}{9} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{142}{9}\right)±\sqrt{\frac{20164}{81}-4\times \frac{49}{9}\left(-\frac{47}{9}\right)}}{2\times \frac{49}{9}}$$

Multiply -4 times \frac{49}{9}.

$$x=\frac{-\left(-\frac{142}{9}\right)±\sqrt{\frac{20164}{81}-\frac{196}{9}\left(-\frac{47}{9}\right)}}{2\times \frac{49}{9}}$$

Multiply -\frac{196}{9} times -\frac{47}{9} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{142}{9}\right)±\sqrt{\frac{20164+9212}{81}}}{2\times \frac{49}{9}}$$

Add \frac{20164}{81} to \frac{9212}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{142}{9}\right)±\sqrt{\frac{1088}{3}}}{2\times \frac{49}{9}}$$

Take the square root of \frac{1088}{3}.

$$x=\frac{-\left(-\frac{142}{9}\right)±\frac{8\sqrt{51}}{3}}{2\times \frac{49}{9}}$$

The opposite of -\frac{142}{9} is \frac{142}{9}.

$$x=\frac{\frac{142}{9}±\frac{8\sqrt{51}}{3}}{2\times \frac{49}{9}}$$

Multiply 2 times \frac{49}{9}.

$$x=\frac{\frac{142}{9}±\frac{8\sqrt{51}}{3}}{\frac{98}{9}}$$

Now solve the equation x=\frac{\frac{142}{9}±\frac{8\sqrt{51}}{3}}{\frac{98}{9}} when ± is plus. Add \frac{142}{9} to \frac{8\sqrt{51}}{3}.

$$x=\frac{\frac{8\sqrt{51}}{3}+\frac{142}{9}}{\frac{98}{9}}$$

Divide \frac{142}{9}+\frac{8\sqrt{51}}{3} by \frac{98}{9} by multiplying \frac{142}{9}+\frac{8\sqrt{51}}{3} by the reciprocal of \frac{98}{9}.

$$x=\frac{12\sqrt{51}+71}{49}$$

Now solve the equation x=\frac{\frac{142}{9}±\frac{8\sqrt{51}}{3}}{\frac{98}{9}} when ± is minus. Subtract \frac{8\sqrt{51}}{3} from \frac{142}{9}.

$$x=\frac{-\frac{8\sqrt{51}}{3}+\frac{142}{9}}{\frac{98}{9}}$$

Divide \frac{142}{9}-\frac{8\sqrt{51}}{3} by \frac{98}{9} by multiplying \frac{142}{9}-\frac{8\sqrt{51}}{3} by the reciprocal of \frac{98}{9}.

$$x=\frac{71-12\sqrt{51}}{49}$$

The equation is now solved.

$$x=\frac{12\sqrt{51}+71}{49} x=\frac{71-12\sqrt{51}}{49}$$

Steps for Completing the Square

To raise \frac{14-7x}{3} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(14-7x\right)^{2}}{3^{2}}+\left(x-3\right)^{2}=\left(6-x\right)^{2}$$

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

$$\frac{\left(14-7x\right)^{2}}{3^{2}}+x^{2}-6x+9=\left(6-x\right)^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{2}-6x+9 times \frac{3^{2}}{3^{2}}.

$$\frac{\left(14-7x\right)^{2}}{3^{2}}+\frac{\left(x^{2}-6x+9\right)\times 3^{2}}{3^{2}}=\left(6-x\right)^{2}$$

Since \frac{\left(14-7x\right)^{2}}{3^{2}} and \frac{\left(x^{2}-6x+9\right)\times 3^{2}}{3^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(14-7x\right)^{2}+\left(x^{2}-6x+9\right)\times 3^{2}}{3^{2}}=\left(6-x\right)^{2}$$

Do the multiplications in \left(14-7x\right)^{2}+\left(x^{2}-6x+9\right)\times 3^{2}.

$$\frac{196-196x+49x^{2}+9x^{2}-54x+81}{3^{2}}=\left(6-x\right)^{2}$$

Combine like terms in 196-196x+49x^{2}+9x^{2}-54x+81.

$$\frac{277-250x+58x^{2}}{3^{2}}=\left(6-x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6-x\right)^{2}.

$$\frac{277-250x+58x^{2}}{3^{2}}=36-12x+x^{2}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{277-250x+58x^{2}}{9}=36-12x+x^{2}$$

Divide each term of 277-250x+58x^{2} by 9 to get \frac{277}{9}-\frac{250}{9}x+\frac{58}{9}x^{2}.

$$\frac{277}{9}-\frac{250}{9}x+\frac{58}{9}x^{2}=36-12x+x^{2}$$

Add 12x to both sides.

$$\frac{277}{9}-\frac{250}{9}x+\frac{58}{9}x^{2}+12x=36+x^{2}$$

Combine -\frac{250}{9}x and 12x to get -\frac{142}{9}x.

$$\frac{277}{9}-\frac{142}{9}x+\frac{58}{9}x^{2}=36+x^{2}$$

Subtract x^{2} from both sides.

$$\frac{277}{9}-\frac{142}{9}x+\frac{58}{9}x^{2}-x^{2}=36$$

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

$$\frac{277}{9}-\frac{142}{9}x+\frac{49}{9}x^{2}=36$$

Subtract \frac{277}{9} from both sides.

$$-\frac{142}{9}x+\frac{49}{9}x^{2}=36-\frac{277}{9}$$

Subtract \frac{277}{9} from 36 to get \frac{47}{9}.

$$-\frac{142}{9}x+\frac{49}{9}x^{2}=\frac{47}{9}$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$\frac{49}{9}x^{2}-\frac{142}{9}x=\frac{47}{9}$$

Divide both sides of the equation by \frac{49}{9}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{49}{9}x^{2}-\frac{142}{9}x}{\frac{49}{9}}=\frac{\frac{47}{9}}{\frac{49}{9}}$$

Dividing by \frac{49}{9} undoes the multiplication by \frac{49}{9}.

$$x^{2}+\left(-\frac{\frac{142}{9}}{\frac{49}{9}}\right)x=\frac{\frac{47}{9}}{\frac{49}{9}}$$

Divide -\frac{142}{9} by \frac{49}{9} by multiplying -\frac{142}{9} by the reciprocal of \frac{49}{9}.

$$x^{2}-\frac{142}{49}x=\frac{\frac{47}{9}}{\frac{49}{9}}$$

Divide \frac{47}{9} by \frac{49}{9} by multiplying \frac{47}{9} by the reciprocal of \frac{49}{9}.

$$x^{2}-\frac{142}{49}x=\frac{47}{49}$$

Divide -\frac{142}{49}, the coefficient of the x term, by 2 to get -\frac{71}{49}. Then add the square of -\frac{71}{49} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{142}{49}x+\left(-\frac{71}{49}\right)^{2}=\frac{47}{49}+\left(-\frac{71}{49}\right)^{2}$$

Square -\frac{71}{49} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{142}{49}x+\frac{5041}{2401}=\frac{47}{49}+\frac{5041}{2401}$$

Add \frac{47}{49} to \frac{5041}{2401} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{142}{49}x+\frac{5041}{2401}=\frac{7344}{2401}$$

Factor x^{2}-\frac{142}{49}x+\frac{5041}{2401}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{71}{49}\right)^{2}=\frac{7344}{2401}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{71}{49}\right)^{2}}=\sqrt{\frac{7344}{2401}}$$

Simplify.

$$x-\frac{71}{49}=\frac{12\sqrt{51}}{49} x-\frac{71}{49}=-\frac{12\sqrt{51}}{49}$$

Add \frac{71}{49} to both sides of the equation.

$$x=\frac{12\sqrt{51}+71}{49} x=\frac{71-12\sqrt{51}}{49}$$