Solve 1/25+1/10+x=x

$$\frac { 1 } { 25 } + \frac { 1 } { 10 + x } = x$$

Solve for x

$$x=\frac{\sqrt{65501}-249}{50}\approx 0.13863263$$

$$x=\frac{-\sqrt{65501}-249}{50}\approx -10.09863263$$

Steps Using the Quadratic Formula

Variable x cannot be equal to -10 since division by zero is not defined. Multiply both sides of the equation by 25\left(x+10\right), the least common multiple of 25,10+x.

$$25\left(x+10\right)\times \frac{1}{25}+25=x\times 25\left(x+10\right)$$

Use the distributive property to multiply 25 by x+10.

$$\left(25x+250\right)\times \frac{1}{25}+25=x\times 25\left(x+10\right)$$

Use the distributive property to multiply 25x+250 by \frac{1}{25}.

$$x+10+25=x\times 25\left(x+10\right)$$

Add 10 and 25 to get 35.

$$x+35=x\times 25\left(x+10\right)$$

Use the distributive property to multiply x\times 25 by x+10.

$$x+35=25x^{2}+10x\times 25$$

Multiply 10 and 25 to get 250.

$$x+35=25x^{2}+250x$$

Subtract 25x^{2} from both sides.

$$x+35-25x^{2}=250x$$

Subtract 250x from both sides.

$$x+35-25x^{2}-250x=0$$

Combine x and -250x to get -249x.

$$-249x+35-25x^{2}=0$$

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.

$$-25x^{2}-249x+35=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -25 for a, -249 for b, and 35 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-249\right)±\sqrt{\left(-249\right)^{2}-4\left(-25\right)\times 35}}{2\left(-25\right)}$$

Square -249.

$$x=\frac{-\left(-249\right)±\sqrt{62001-4\left(-25\right)\times 35}}{2\left(-25\right)}$$

Multiply -4 times -25.

$$x=\frac{-\left(-249\right)±\sqrt{62001+100\times 35}}{2\left(-25\right)}$$

Multiply 100 times 35.

$$x=\frac{-\left(-249\right)±\sqrt{62001+3500}}{2\left(-25\right)}$$

Add 62001 to 3500.

$$x=\frac{-\left(-249\right)±\sqrt{65501}}{2\left(-25\right)}$$

The opposite of -249 is 249.

$$x=\frac{249±\sqrt{65501}}{2\left(-25\right)}$$

Multiply 2 times -25.

$$x=\frac{249±\sqrt{65501}}{-50}$$

Now solve the equation x=\frac{249±\sqrt{65501}}{-50} when ± is plus. Add 249 to \sqrt{65501}.

$$x=\frac{\sqrt{65501}+249}{-50}$$

Divide 249+\sqrt{65501} by -50.

$$x=\frac{-\sqrt{65501}-249}{50}$$

Now solve the equation x=\frac{249±\sqrt{65501}}{-50} when ± is minus. Subtract \sqrt{65501} from 249.

$$x=\frac{249-\sqrt{65501}}{-50}$$

Divide 249-\sqrt{65501} by -50.

$$x=\frac{\sqrt{65501}-249}{50}$$

The equation is now solved.

$$x=\frac{-\sqrt{65501}-249}{50} x=\frac{\sqrt{65501}-249}{50}$$

Steps for Completing the Square

Variable x cannot be equal to -10 since division by zero is not defined. Multiply both sides of the equation by 25\left(x+10\right), the least common multiple of 25,10+x.

$$25\left(x+10\right)\times \frac{1}{25}+25=x\times 25\left(x+10\right)$$

Use the distributive property to multiply 25 by x+10.

$$\left(25x+250\right)\times \frac{1}{25}+25=x\times 25\left(x+10\right)$$

Use the distributive property to multiply 25x+250 by \frac{1}{25}.

$$x+10+25=x\times 25\left(x+10\right)$$

Add 10 and 25 to get 35.

$$x+35=x\times 25\left(x+10\right)$$

Use the distributive property to multiply x\times 25 by x+10.

$$x+35=25x^{2}+10x\times 25$$

Multiply 10 and 25 to get 250.

$$x+35=25x^{2}+250x$$

Subtract 25x^{2} from both sides.

$$x+35-25x^{2}=250x$$

Subtract 250x from both sides.

$$x+35-25x^{2}-250x=0$$

Combine x and -250x to get -249x.

$$-249x+35-25x^{2}=0$$

Subtract 35 from both sides. Anything subtracted from zero gives its negation.

$$-249x-25x^{2}=-35$$

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.

$$-25x^{2}-249x=-35$$

Divide both sides by -25.

$$\frac{-25x^{2}-249x}{-25}=-\frac{35}{-25}$$

Dividing by -25 undoes the multiplication by -25.

$$x^{2}+\left(-\frac{249}{-25}\right)x=-\frac{35}{-25}$$

Divide -249 by -25.

$$x^{2}+\frac{249}{25}x=-\frac{35}{-25}$$

Reduce the fraction \frac{-35}{-25} to lowest terms by extracting and canceling out 5.

$$x^{2}+\frac{249}{25}x=\frac{7}{5}$$

Divide \frac{249}{25}, the coefficient of the x term, by 2 to get \frac{249}{50}. Then add the square of \frac{249}{50} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{249}{25}x+\left(\frac{249}{50}\right)^{2}=\frac{7}{5}+\left(\frac{249}{50}\right)^{2}$$

Square \frac{249}{50} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{249}{25}x+\frac{62001}{2500}=\frac{7}{5}+\frac{62001}{2500}$$

Add \frac{7}{5} to \frac{62001}{2500} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{249}{25}x+\frac{62001}{2500}=\frac{65501}{2500}$$

Factor x^{2}+\frac{249}{25}x+\frac{62001}{2500}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{249}{50}\right)^{2}=\frac{65501}{2500}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{249}{50}\right)^{2}}=\sqrt{\frac{65501}{2500}}$$

Simplify.

$$x+\frac{249}{50}=\frac{\sqrt{65501}}{50} x+\frac{249}{50}=-\frac{\sqrt{65501}}{50}$$

Subtract \frac{249}{50} from both sides of the equation.

$$x=\frac{\sqrt{65501}-249}{50} x=\frac{-\sqrt{65501}-249}{50}$$