Solve {left(65-5x/13right)}^2+x^2=25

$${ \left( \frac{ 65-5x }{ 13 } \right) }^{ 2 } + { x }^{ 2 } =25$$

Solve for x

$$x = \frac{325}{97} = 3\frac{34}{97} \approx 3.350515464$$

$$x=0$$

Steps Using Factoring

To raise \frac{65-5x}{13} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(65-5x\right)^{2}}{13^{2}}+x^{2}=25$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{2} times \frac{13^{2}}{13^{2}}.

$$\frac{\left(65-5x\right)^{2}}{13^{2}}+\frac{x^{2}\times 13^{2}}{13^{2}}=25$$

Since \frac{\left(65-5x\right)^{2}}{13^{2}} and \frac{x^{2}\times 13^{2}}{13^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(65-5x\right)^{2}+x^{2}\times 13^{2}}{13^{2}}=25$$

Do the multiplications in \left(65-5x\right)^{2}+x^{2}\times 13^{2}.

$$\frac{4225-650x+25x^{2}+169x^{2}}{13^{2}}=25$$

Combine like terms in 4225-650x+25x^{2}+169x^{2}.

$$\frac{4225-650x+194x^{2}}{13^{2}}=25$$

Calculate 13 to the power of 2 and get 169.

$$\frac{4225-650x+194x^{2}}{169}=25$$

Divide each term of 4225-650x+194x^{2} by 169 to get 25-\frac{50}{13}x+\frac{194}{169}x^{2}.

$$25-\frac{50}{13}x+\frac{194}{169}x^{2}=25$$

Subtract 25 from both sides.

$$25-\frac{50}{13}x+\frac{194}{169}x^{2}-25=0$$

Subtract 25 from 25 to get 0.

$$-\frac{50}{13}x+\frac{194}{169}x^{2}=0$$

Factor out x.

$$x\left(-\frac{50}{13}+\frac{194}{169}x\right)=0$$

To find equation solutions, solve x=0 and -\frac{50}{13}+\frac{194x}{169}=0.

$$x=0 x=\frac{325}{97}$$

Steps Using the Quadratic Formula

To raise \frac{65-5x}{13} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(65-5x\right)^{2}}{13^{2}}+x^{2}=25$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{2} times \frac{13^{2}}{13^{2}}.

$$\frac{\left(65-5x\right)^{2}}{13^{2}}+\frac{x^{2}\times 13^{2}}{13^{2}}=25$$

Since \frac{\left(65-5x\right)^{2}}{13^{2}} and \frac{x^{2}\times 13^{2}}{13^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(65-5x\right)^{2}+x^{2}\times 13^{2}}{13^{2}}=25$$

Do the multiplications in \left(65-5x\right)^{2}+x^{2}\times 13^{2}.

$$\frac{4225-650x+25x^{2}+169x^{2}}{13^{2}}=25$$

Combine like terms in 4225-650x+25x^{2}+169x^{2}.

$$\frac{4225-650x+194x^{2}}{13^{2}}=25$$

Calculate 13 to the power of 2 and get 169.

$$\frac{4225-650x+194x^{2}}{169}=25$$

Divide each term of 4225-650x+194x^{2} by 169 to get 25-\frac{50}{13}x+\frac{194}{169}x^{2}.

$$25-\frac{50}{13}x+\frac{194}{169}x^{2}=25$$

Subtract 25 from both sides.

$$25-\frac{50}{13}x+\frac{194}{169}x^{2}-25=0$$

Subtract 25 from 25 to get 0.

$$-\frac{50}{13}x+\frac{194}{169}x^{2}=0$$

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.

$$\frac{194}{169}x^{2}-\frac{50}{13}x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{194}{169} for a, -\frac{50}{13} for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{50}{13}\right)±\sqrt{\left(-\frac{50}{13}\right)^{2}}}{2\times \frac{194}{169}}$$

Take the square root of \left(-\frac{50}{13}\right)^{2}.

$$x=\frac{-\left(-\frac{50}{13}\right)±\frac{50}{13}}{2\times \frac{194}{169}}$$

The opposite of -\frac{50}{13} is \frac{50}{13}.

$$x=\frac{\frac{50}{13}±\frac{50}{13}}{2\times \frac{194}{169}}$$

Multiply 2 times \frac{194}{169}.

$$x=\frac{\frac{50}{13}±\frac{50}{13}}{\frac{388}{169}}$$

Now solve the equation x=\frac{\frac{50}{13}±\frac{50}{13}}{\frac{388}{169}} when ± is plus. Add \frac{50}{13} to \frac{50}{13} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{\frac{100}{13}}{\frac{388}{169}}$$

Divide \frac{100}{13} by \frac{388}{169} by multiplying \frac{100}{13} by the reciprocal of \frac{388}{169}.

$$x=\frac{325}{97}$$

Now solve the equation x=\frac{\frac{50}{13}±\frac{50}{13}}{\frac{388}{169}} when ± is minus. Subtract \frac{50}{13} from \frac{50}{13} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{0}{\frac{388}{169}}$$

Divide 0 by \frac{388}{169} by multiplying 0 by the reciprocal of \frac{388}{169}.

$$x=0$$

The equation is now solved.

$$x=\frac{325}{97} x=0$$

Steps for Completing the Square

To raise \frac{65-5x}{13} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(65-5x\right)^{2}}{13^{2}}+x^{2}=25$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{2} times \frac{13^{2}}{13^{2}}.

$$\frac{\left(65-5x\right)^{2}}{13^{2}}+\frac{x^{2}\times 13^{2}}{13^{2}}=25$$

Since \frac{\left(65-5x\right)^{2}}{13^{2}} and \frac{x^{2}\times 13^{2}}{13^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(65-5x\right)^{2}+x^{2}\times 13^{2}}{13^{2}}=25$$

Do the multiplications in \left(65-5x\right)^{2}+x^{2}\times 13^{2}.

$$\frac{4225-650x+25x^{2}+169x^{2}}{13^{2}}=25$$

Combine like terms in 4225-650x+25x^{2}+169x^{2}.

$$\frac{4225-650x+194x^{2}}{13^{2}}=25$$

Calculate 13 to the power of 2 and get 169.

$$\frac{4225-650x+194x^{2}}{169}=25$$

Divide each term of 4225-650x+194x^{2} by 169 to get 25-\frac{50}{13}x+\frac{194}{169}x^{2}.

$$25-\frac{50}{13}x+\frac{194}{169}x^{2}=25$$

Subtract 25 from both sides.

$$-\frac{50}{13}x+\frac{194}{169}x^{2}=25-25$$

Subtract 25 from 25 to get 0.

$$-\frac{50}{13}x+\frac{194}{169}x^{2}=0$$

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.

$$\frac{194}{169}x^{2}-\frac{50}{13}x=0$$

Divide both sides of the equation by \frac{194}{169}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{194}{169}x^{2}-\frac{50}{13}x}{\frac{194}{169}}=\frac{0}{\frac{194}{169}}$$

Dividing by \frac{194}{169} undoes the multiplication by \frac{194}{169}.

$$x^{2}+\left(-\frac{\frac{50}{13}}{\frac{194}{169}}\right)x=\frac{0}{\frac{194}{169}}$$

Divide -\frac{50}{13} by \frac{194}{169} by multiplying -\frac{50}{13} by the reciprocal of \frac{194}{169}.

$$x^{2}-\frac{325}{97}x=\frac{0}{\frac{194}{169}}$$

Divide 0 by \frac{194}{169} by multiplying 0 by the reciprocal of \frac{194}{169}.

$$x^{2}-\frac{325}{97}x=0$$

Divide -\frac{325}{97}, the coefficient of the x term, by 2 to get -\frac{325}{194}. Then add the square of -\frac{325}{194} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{325}{97}x+\left(-\frac{325}{194}\right)^{2}=\left(-\frac{325}{194}\right)^{2}$$

Square -\frac{325}{194} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{325}{97}x+\frac{105625}{37636}=\frac{105625}{37636}$$

Factor x^{2}-\frac{325}{97}x+\frac{105625}{37636}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{325}{194}\right)^{2}=\frac{105625}{37636}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{325}{194}\right)^{2}}=\sqrt{\frac{105625}{37636}}$$

Simplify.

$$x-\frac{325}{194}=\frac{325}{194} x-\frac{325}{194}=-\frac{325}{194}$$

Add \frac{325}{194} to both sides of the equation.

$$x=\frac{325}{97} x=0$$