Solve [-50x+5]^2/5+10+50x^2=25

$$\frac { [ - 50 x + 5 ] ^ { 2 } } { 5 + 10 + 50 x ^ { 2 } } = 25$$

Solve for x

$$x=\frac{2\sqrt{2}+1}{5}\approx 0.765685425$$

$$x=\frac{1-2\sqrt{2}}{5}\approx -0.365685425$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 5\left(10x^{2}+3\right).

$$\left(-50x+5\right)^{2}=125\left(10x^{2}+3\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-50x+5\right)^{2}.

$$2500x^{2}-500x+25=125\left(10x^{2}+3\right)$$

Use the distributive property to multiply 125 by 10x^{2}+3.

$$2500x^{2}-500x+25=1250x^{2}+375$$

Subtract 1250x^{2} from both sides.

$$2500x^{2}-500x+25-1250x^{2}=375$$

Combine 2500x^{2} and -1250x^{2} to get 1250x^{2}.

$$1250x^{2}-500x+25=375$$

Subtract 375 from both sides.

$$1250x^{2}-500x+25-375=0$$

Subtract 375 from 25 to get -350.

$$1250x^{2}-500x-350=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1250 for a, -500 for b, and -350 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-500\right)±\sqrt{\left(-500\right)^{2}-4\times 1250\left(-350\right)}}{2\times 1250}$$

Square -500.

$$x=\frac{-\left(-500\right)±\sqrt{250000-4\times 1250\left(-350\right)}}{2\times 1250}$$

Multiply -4 times 1250.

$$x=\frac{-\left(-500\right)±\sqrt{250000-5000\left(-350\right)}}{2\times 1250}$$

Multiply -5000 times -350.

$$x=\frac{-\left(-500\right)±\sqrt{250000+1750000}}{2\times 1250}$$

Add 250000 to 1750000.

$$x=\frac{-\left(-500\right)±\sqrt{2000000}}{2\times 1250}$$

Take the square root of 2000000.

$$x=\frac{-\left(-500\right)±1000\sqrt{2}}{2\times 1250}$$

The opposite of -500 is 500.

$$x=\frac{500±1000\sqrt{2}}{2\times 1250}$$

Multiply 2 times 1250.

$$x=\frac{500±1000\sqrt{2}}{2500}$$

Now solve the equation x=\frac{500±1000\sqrt{2}}{2500} when ± is plus. Add 500 to 1000\sqrt{2}.

$$x=\frac{1000\sqrt{2}+500}{2500}$$

Divide 1000\sqrt{2}+500 by 2500.

$$x=\frac{2\sqrt{2}+1}{5}$$

Now solve the equation x=\frac{500±1000\sqrt{2}}{2500} when ± is minus. Subtract 1000\sqrt{2} from 500.

$$x=\frac{500-1000\sqrt{2}}{2500}$$

Divide 500-1000\sqrt{2} by 2500.

$$x=\frac{1-2\sqrt{2}}{5}$$

The equation is now solved.

$$x=\frac{2\sqrt{2}+1}{5} x=\frac{1-2\sqrt{2}}{5}$$

Steps for Completing the Square

Multiply both sides of the equation by 5\left(10x^{2}+3\right).

$$\left(-50x+5\right)^{2}=125\left(10x^{2}+3\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-50x+5\right)^{2}.

$$2500x^{2}-500x+25=125\left(10x^{2}+3\right)$$

Use the distributive property to multiply 125 by 10x^{2}+3.

$$2500x^{2}-500x+25=1250x^{2}+375$$

Subtract 1250x^{2} from both sides.

$$2500x^{2}-500x+25-1250x^{2}=375$$

Combine 2500x^{2} and -1250x^{2} to get 1250x^{2}.

$$1250x^{2}-500x+25=375$$

Subtract 25 from both sides.

$$1250x^{2}-500x=375-25$$

Subtract 25 from 375 to get 350.

$$1250x^{2}-500x=350$$

Divide both sides by 1250.

$$\frac{1250x^{2}-500x}{1250}=\frac{350}{1250}$$

Dividing by 1250 undoes the multiplication by 1250.

$$x^{2}+\left(-\frac{500}{1250}\right)x=\frac{350}{1250}$$

Reduce the fraction \frac{-500}{1250} to lowest terms by extracting and canceling out 250.

$$x^{2}-\frac{2}{5}x=\frac{350}{1250}$$

Reduce the fraction \frac{350}{1250} to lowest terms by extracting and canceling out 50.

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

Divide -\frac{2}{5}, the coefficient of the x term, by 2 to get -\frac{1}{5}. Then add the square of -\frac{1}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{2}{5}x+\left(-\frac{1}{5}\right)^{2}=\frac{7}{25}+\left(-\frac{1}{5}\right)^{2}$$

Square -\frac{1}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{2}{5}x+\frac{1}{25}=\frac{7+1}{25}$$

Add \frac{7}{25} to \frac{1}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{2}{5}x+\frac{1}{25}=\frac{8}{25}$$

Factor x^{2}-\frac{2}{5}x+\frac{1}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{5}\right)^{2}}=\sqrt{\frac{8}{25}}$$

Simplify.

$$x-\frac{1}{5}=\frac{2\sqrt{2}}{5} x-\frac{1}{5}=-\frac{2\sqrt{2}}{5}$$

Add \frac{1}{5} to both sides of the equation.

$$x=\frac{2\sqrt{2}+1}{5} x=\frac{1-2\sqrt{2}}{5}$$