Solve (5sqrt{5}-x)^2+25=x

$$( 5 \sqrt { 5 } - x ) ^ { 2 } + 25 = x$$

Solve for x (complex solution)

$$x=\frac{i\sqrt{99-20\sqrt{5}}}{2}+5\sqrt{5}+\frac{1}{2}\approx 11.680339887+3.683701958i$$

$$x=-\frac{i\sqrt{99-20\sqrt{5}}}{2}+5\sqrt{5}+\frac{1}{2}\approx 11.680339887-3.683701958i$$

Steps Using the Quadratic Formula

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

$$25\left(\sqrt{5}\right)^{2}-10\sqrt{5}x+x^{2}+25=x$$

The square of \sqrt{5} is 5.

$$25\times 5-10\sqrt{5}x+x^{2}+25=x$$

Multiply 25 and 5 to get 125.

$$125-10\sqrt{5}x+x^{2}+25=x$$

Add 125 and 25 to get 150.

$$150-10\sqrt{5}x+x^{2}=x$$

Subtract x from both sides.

$$150-10\sqrt{5}x+x^{2}-x=0$$

Combine all terms containing x.

$$150+\left(-10\sqrt{5}-1\right)x+x^{2}=0$$

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.

$$x^{2}+\left(-10\sqrt{5}-1\right)x+150=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10\sqrt{5}-1 for b, and 150 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\sqrt{5}-1\right)±\sqrt{\left(-10\sqrt{5}-1\right)^{2}-4\times 150}}{2}$$

Square -10\sqrt{5}-1.

$$x=\frac{-\left(-10\sqrt{5}-1\right)±\sqrt{20\sqrt{5}+501-4\times 150}}{2}$$

Multiply -4 times 150.

$$x=\frac{-\left(-10\sqrt{5}-1\right)±\sqrt{20\sqrt{5}+501-600}}{2}$$

Add 501+20\sqrt{5} to -600.

$$x=\frac{-\left(-10\sqrt{5}-1\right)±\sqrt{20\sqrt{5}-99}}{2}$$

Take the square root of -99+20\sqrt{5}.

$$x=\frac{-\left(-10\sqrt{5}-1\right)±i\sqrt{99-20\sqrt{5}}}{2}$$

The opposite of -10\sqrt{5}-1 is 10\sqrt{5}+1.

$$x=\frac{10\sqrt{5}+1±i\sqrt{99-20\sqrt{5}}}{2}$$

Now solve the equation x=\frac{10\sqrt{5}+1±i\sqrt{99-20\sqrt{5}}}{2} when ± is plus. Add 10\sqrt{5}+1 to i\sqrt{99-20\sqrt{5}}.

$$x=\frac{10\sqrt{5}+1+i\sqrt{99-20\sqrt{5}}}{2}$$

Divide 10\sqrt{5}+1+i\sqrt{99-20\sqrt{5}} by 2.

$$x=\frac{i\sqrt{99-20\sqrt{5}}}{2}+5\sqrt{5}+\frac{1}{2}$$

Now solve the equation x=\frac{10\sqrt{5}+1±i\sqrt{99-20\sqrt{5}}}{2} when ± is minus. Subtract i\sqrt{99-20\sqrt{5}} from 10\sqrt{5}+1.

$$x=\frac{-i\sqrt{99-20\sqrt{5}}+10\sqrt{5}+1}{2}$$

Divide 10\sqrt{5}+1-i\sqrt{99-20\sqrt{5}} by 2.

$$x=-\frac{i\sqrt{99-20\sqrt{5}}}{2}+5\sqrt{5}+\frac{1}{2}$$

The equation is now solved.

$$x=\frac{i\sqrt{99-20\sqrt{5}}}{2}+5\sqrt{5}+\frac{1}{2} x=-\frac{i\sqrt{99-20\sqrt{5}}}{2}+5\sqrt{5}+\frac{1}{2}$$

Steps for Completing the Square

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

$$25\left(\sqrt{5}\right)^{2}-10\sqrt{5}x+x^{2}+25=x$$

The square of \sqrt{5} is 5.

$$25\times 5-10\sqrt{5}x+x^{2}+25=x$$

Multiply 25 and 5 to get 125.

$$125-10\sqrt{5}x+x^{2}+25=x$$

Add 125 and 25 to get 150.

$$150-10\sqrt{5}x+x^{2}=x$$

Subtract x from both sides.

$$150-10\sqrt{5}x+x^{2}-x=0$$

Subtract 150 from both sides. Anything subtracted from zero gives its negation.

$$-10\sqrt{5}x+x^{2}-x=-150$$

Combine all terms containing x.

$$\left(-10\sqrt{5}-1\right)x+x^{2}=-150$$

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.

$$x^{2}+\left(-10\sqrt{5}-1\right)x=-150$$

Divide -10\sqrt{5}-1, the coefficient of the x term, by 2 to get -5\sqrt{5}-\frac{1}{2}. Then add the square of -5\sqrt{5}-\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\left(-10\sqrt{5}-1\right)x+\left(-5\sqrt{5}-\frac{1}{2}\right)^{2}=-150+\left(-5\sqrt{5}-\frac{1}{2}\right)^{2}$$

Square -5\sqrt{5}-\frac{1}{2}.

$$x^{2}+\left(-10\sqrt{5}-1\right)x+5\sqrt{5}+\frac{501}{4}=-150+5\sqrt{5}+\frac{501}{4}$$

Add -150 to \frac{501}{4}+5\sqrt{5}.

$$x^{2}+\left(-10\sqrt{5}-1\right)x+5\sqrt{5}+\frac{501}{4}=5\sqrt{5}-\frac{99}{4}$$

Factor x^{2}+\left(-10\sqrt{5}-1\right)x+5\sqrt{5}+\frac{501}{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}.

$$\left(x-5\sqrt{5}-\frac{1}{2}\right)^{2}=5\sqrt{5}-\frac{99}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\sqrt{5}-\frac{1}{2}\right)^{2}}=\sqrt{5\sqrt{5}-\frac{99}{4}}$$

Simplify.

$$x-5\sqrt{5}-\frac{1}{2}=\frac{i\sqrt{99-20\sqrt{5}}}{2} x-5\sqrt{5}-\frac{1}{2}=-\frac{i\sqrt{99-20\sqrt{5}}}{2}$$

Subtract -5\sqrt{5}-\frac{1}{2} from both sides of the equation.

$$x=\frac{i\sqrt{99-20\sqrt{5}}}{2}+5\sqrt{5}+\frac{1}{2} x=-\frac{i\sqrt{99-20\sqrt{5}}}{2}+5\sqrt{5}+\frac{1}{2}$$