Solve (x-25/4)=-1/25x^2

$$(x- \frac{ 25 }{ 4 } )=- \frac{ 1 }{ 25 } { x }^{ 2 }$$

Solve for x

$$x = \frac{25 \sqrt{2} - 25}{2} \approx 5.17766953$$

$$x=\frac{-25\sqrt{2}-25}{2}\approx -30.17766953$$

Steps Using the Quadratic Formula

Add \frac{1}{25}x^{2} to both sides.

$$x-\frac{25}{4}+\frac{1}{25}x^{2}=0$$

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.

$$\frac{1}{25}x^{2}+x-\frac{25}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{25} for a, 1 for b, and -\frac{25}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\times \frac{1}{25}\left(-\frac{25}{4}\right)}}{2\times \frac{1}{25}}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times \frac{1}{25}\left(-\frac{25}{4}\right)}}{2\times \frac{1}{25}}$$

Multiply -4 times \frac{1}{25}.

$$x=\frac{-1±\sqrt{1-\frac{4}{25}\left(-\frac{25}{4}\right)}}{2\times \frac{1}{25}}$$

Multiply -\frac{4}{25} times -\frac{25}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-1±\sqrt{1+1}}{2\times \frac{1}{25}}$$

Add 1 to 1.

$$x=\frac{-1±\sqrt{2}}{2\times \frac{1}{25}}$$

Multiply 2 times \frac{1}{25}.

$$x=\frac{-1±\sqrt{2}}{\frac{2}{25}}$$

Now solve the equation x=\frac{-1±\sqrt{2}}{\frac{2}{25}} when ± is plus. Add -1 to \sqrt{2}.

$$x=\frac{\sqrt{2}-1}{\frac{2}{25}}$$

Divide -1+\sqrt{2} by \frac{2}{25} by multiplying -1+\sqrt{2} by the reciprocal of \frac{2}{25}.

$$x=\frac{25\sqrt{2}-25}{2}$$

Now solve the equation x=\frac{-1±\sqrt{2}}{\frac{2}{25}} when ± is minus. Subtract \sqrt{2} from -1.

$$x=\frac{-\sqrt{2}-1}{\frac{2}{25}}$$

Divide -1-\sqrt{2} by \frac{2}{25} by multiplying -1-\sqrt{2} by the reciprocal of \frac{2}{25}.

$$x=\frac{-25\sqrt{2}-25}{2}$$

The equation is now solved.

$$x=\frac{25\sqrt{2}-25}{2} x=\frac{-25\sqrt{2}-25}{2}$$

Steps for Completing the Square

Add \frac{1}{25}x^{2} to both sides.

$$x-\frac{25}{4}+\frac{1}{25}x^{2}=0$$

Add \frac{25}{4} to both sides. Anything plus zero gives itself.

$$x+\frac{1}{25}x^{2}=\frac{25}{4}$$

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.

$$\frac{1}{25}x^{2}+x=\frac{25}{4}$$

Multiply both sides by 25.

$$\frac{\frac{1}{25}x^{2}+x}{\frac{1}{25}}=\frac{\frac{25}{4}}{\frac{1}{25}}$$

Dividing by \frac{1}{25} undoes the multiplication by \frac{1}{25}.

$$x^{2}+\frac{1}{\frac{1}{25}}x=\frac{\frac{25}{4}}{\frac{1}{25}}$$

Divide 1 by \frac{1}{25} by multiplying 1 by the reciprocal of \frac{1}{25}.

$$x^{2}+25x=\frac{\frac{25}{4}}{\frac{1}{25}}$$

Divide \frac{25}{4} by \frac{1}{25} by multiplying \frac{25}{4} by the reciprocal of \frac{1}{25}.

$$x^{2}+25x=\frac{625}{4}$$

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

$$x^{2}+25x+\left(\frac{25}{2}\right)^{2}=\frac{625}{4}+\left(\frac{25}{2}\right)^{2}$$

Square \frac{25}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+25x+\frac{625}{4}=\frac{625+625}{4}$$

Add \frac{625}{4} to \frac{625}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+25x+\frac{625}{4}=\frac{625}{2}$$

Factor x^{2}+25x+\frac{625}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{25}{2}\right)^{2}=\frac{625}{2}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{25}{2}\right)^{2}}=\sqrt{\frac{625}{2}}$$

Simplify.

$$x+\frac{25}{2}=\frac{25\sqrt{2}}{2} x+\frac{25}{2}=-\frac{25\sqrt{2}}{2}$$

Subtract \frac{25}{2} from both sides of the equation.

$$x=\frac{25\sqrt{2}-25}{2} x=\frac{-25\sqrt{2}-25}{2}$$