Solve 1/16x^2+4/25(1-x)^2-2/5x(1-x)=0

$$\frac { 1 } { 16 } x ^ { 2 } + \frac { 4 } { 25 } ( 1 - x ) ^ { 2 } - \frac { 2 } { 5 } x ( 1 - x ) = 0$$

Solve for x

$$x=\frac{40\sqrt{3}}{249}+\frac{48}{83}\approx 0.856554347$$

$$x=-\frac{40\sqrt{3}}{249}+\frac{48}{83}\approx 0.300072159$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-x\right)^{2}.

$$\frac{1}{16}x^{2}+\frac{4}{25}\left(1-2x+x^{2}\right)-\frac{2}{5}x\left(1-x\right)=0$$

Use the distributive property to multiply \frac{4}{25} by 1-2x+x^{2}.

$$\frac{1}{16}x^{2}+\frac{4}{25}-\frac{8}{25}x+\frac{4}{25}x^{2}-\frac{2}{5}x\left(1-x\right)=0$$

Combine \frac{1}{16}x^{2} and \frac{4}{25}x^{2} to get \frac{89}{400}x^{2}.

$$\frac{89}{400}x^{2}+\frac{4}{25}-\frac{8}{25}x-\frac{2}{5}x\left(1-x\right)=0$$

Use the distributive property to multiply -\frac{2}{5}x by 1-x.

$$\frac{89}{400}x^{2}+\frac{4}{25}-\frac{8}{25}x-\frac{2}{5}x+\frac{2}{5}x^{2}=0$$

Combine -\frac{8}{25}x and -\frac{2}{5}x to get -\frac{18}{25}x.

$$\frac{89}{400}x^{2}+\frac{4}{25}-\frac{18}{25}x+\frac{2}{5}x^{2}=0$$

Combine \frac{89}{400}x^{2} and \frac{2}{5}x^{2} to get \frac{249}{400}x^{2}.

$$\frac{249}{400}x^{2}+\frac{4}{25}-\frac{18}{25}x=0$$

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.

$$\frac{249}{400}x^{2}-\frac{18}{25}x+\frac{4}{25}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{249}{400} for a, -\frac{18}{25} for b, and \frac{4}{25} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{18}{25}\right)±\sqrt{\left(-\frac{18}{25}\right)^{2}-4\times \frac{249}{400}\times \frac{4}{25}}}{2\times \frac{249}{400}}$$

Square -\frac{18}{25} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{18}{25}\right)±\sqrt{\frac{324}{625}-4\times \frac{249}{400}\times \frac{4}{25}}}{2\times \frac{249}{400}}$$

Multiply -4 times \frac{249}{400}.

$$x=\frac{-\left(-\frac{18}{25}\right)±\sqrt{\frac{324}{625}-\frac{249}{100}\times \frac{4}{25}}}{2\times \frac{249}{400}}$$

Multiply -\frac{249}{100} times \frac{4}{25} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{18}{25}\right)±\sqrt{\frac{324-249}{625}}}{2\times \frac{249}{400}}$$

Add \frac{324}{625} to -\frac{249}{625} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{18}{25}\right)±\sqrt{\frac{3}{25}}}{2\times \frac{249}{400}}$$

Take the square root of \frac{3}{25}.

$$x=\frac{-\left(-\frac{18}{25}\right)±\frac{\sqrt{3}}{5}}{2\times \frac{249}{400}}$$

The opposite of -\frac{18}{25} is \frac{18}{25}.

$$x=\frac{\frac{18}{25}±\frac{\sqrt{3}}{5}}{2\times \frac{249}{400}}$$

Multiply 2 times \frac{249}{400}.

$$x=\frac{\frac{18}{25}±\frac{\sqrt{3}}{5}}{\frac{249}{200}}$$

Now solve the equation x=\frac{\frac{18}{25}±\frac{\sqrt{3}}{5}}{\frac{249}{200}} when ± is plus. Add \frac{18}{25} to \frac{\sqrt{3}}{5}.

$$x=\frac{\frac{\sqrt{3}}{5}+\frac{18}{25}}{\frac{249}{200}}$$

Divide \frac{18}{25}+\frac{\sqrt{3}}{5} by \frac{249}{200} by multiplying \frac{18}{25}+\frac{\sqrt{3}}{5} by the reciprocal of \frac{249}{200}.

$$x=\frac{40\sqrt{3}}{249}+\frac{48}{83}$$

Now solve the equation x=\frac{\frac{18}{25}±\frac{\sqrt{3}}{5}}{\frac{249}{200}} when ± is minus. Subtract \frac{\sqrt{3}}{5} from \frac{18}{25}.

$$x=\frac{-\frac{\sqrt{3}}{5}+\frac{18}{25}}{\frac{249}{200}}$$

Divide \frac{18}{25}-\frac{\sqrt{3}}{5} by \frac{249}{200} by multiplying \frac{18}{25}-\frac{\sqrt{3}}{5} by the reciprocal of \frac{249}{200}.

$$x=-\frac{40\sqrt{3}}{249}+\frac{48}{83}$$

The equation is now solved.

$$x=\frac{40\sqrt{3}}{249}+\frac{48}{83} x=-\frac{40\sqrt{3}}{249}+\frac{48}{83}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-x\right)^{2}.

$$\frac{1}{16}x^{2}+\frac{4}{25}\left(1-2x+x^{2}\right)-\frac{2}{5}x\left(1-x\right)=0$$

Use the distributive property to multiply \frac{4}{25} by 1-2x+x^{2}.

$$\frac{1}{16}x^{2}+\frac{4}{25}-\frac{8}{25}x+\frac{4}{25}x^{2}-\frac{2}{5}x\left(1-x\right)=0$$

Combine \frac{1}{16}x^{2} and \frac{4}{25}x^{2} to get \frac{89}{400}x^{2}.

$$\frac{89}{400}x^{2}+\frac{4}{25}-\frac{8}{25}x-\frac{2}{5}x\left(1-x\right)=0$$

Use the distributive property to multiply -\frac{2}{5}x by 1-x.

$$\frac{89}{400}x^{2}+\frac{4}{25}-\frac{8}{25}x-\frac{2}{5}x+\frac{2}{5}x^{2}=0$$

Combine -\frac{8}{25}x and -\frac{2}{5}x to get -\frac{18}{25}x.

$$\frac{89}{400}x^{2}+\frac{4}{25}-\frac{18}{25}x+\frac{2}{5}x^{2}=0$$

Combine \frac{89}{400}x^{2} and \frac{2}{5}x^{2} to get \frac{249}{400}x^{2}.

$$\frac{249}{400}x^{2}+\frac{4}{25}-\frac{18}{25}x=0$$

Subtract \frac{4}{25} from both sides. Anything subtracted from zero gives its negation.

$$\frac{249}{400}x^{2}-\frac{18}{25}x=-\frac{4}{25}$$

Divide both sides of the equation by \frac{249}{400}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{249}{400}x^{2}-\frac{18}{25}x}{\frac{249}{400}}=-\frac{\frac{4}{25}}{\frac{249}{400}}$$

Dividing by \frac{249}{400} undoes the multiplication by \frac{249}{400}.

$$x^{2}+\left(-\frac{\frac{18}{25}}{\frac{249}{400}}\right)x=-\frac{\frac{4}{25}}{\frac{249}{400}}$$

Divide -\frac{18}{25} by \frac{249}{400} by multiplying -\frac{18}{25} by the reciprocal of \frac{249}{400}.

$$x^{2}-\frac{96}{83}x=-\frac{\frac{4}{25}}{\frac{249}{400}}$$

Divide -\frac{4}{25} by \frac{249}{400} by multiplying -\frac{4}{25} by the reciprocal of \frac{249}{400}.

$$x^{2}-\frac{96}{83}x=-\frac{64}{249}$$

Divide -\frac{96}{83}, the coefficient of the x term, by 2 to get -\frac{48}{83}. Then add the square of -\frac{48}{83} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{96}{83}x+\left(-\frac{48}{83}\right)^{2}=-\frac{64}{249}+\left(-\frac{48}{83}\right)^{2}$$

Square -\frac{48}{83} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{96}{83}x+\frac{2304}{6889}=-\frac{64}{249}+\frac{2304}{6889}$$

Add -\frac{64}{249} to \frac{2304}{6889} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{96}{83}x+\frac{2304}{6889}=\frac{1600}{20667}$$

Factor x^{2}-\frac{96}{83}x+\frac{2304}{6889}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{48}{83}\right)^{2}=\frac{1600}{20667}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{48}{83}\right)^{2}}=\sqrt{\frac{1600}{20667}}$$

Simplify.

$$x-\frac{48}{83}=\frac{40\sqrt{3}}{249} x-\frac{48}{83}=-\frac{40\sqrt{3}}{249}$$

Add \frac{48}{83} to both sides of the equation.

$$x=\frac{40\sqrt{3}}{249}+\frac{48}{83} x=-\frac{40\sqrt{3}}{249}+\frac{48}{83}$$