Solve =(4x)^2+x^2=(18-x)^2

$$= ( 4 x ) ^ { 2 } + x ^ { 2 } = ( 18 - x ) ^ { 2 }$$

Solve for x

$$x = \frac{9 \sqrt{17} - 9}{8} \approx 3.513493829$$

$$x=\frac{-9\sqrt{17}-9}{8}\approx -5.763493829$$

Steps Using the Quadratic Formula

Expand \left(4x\right)^{2}.

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

Calculate 4 to the power of 2 and get 16.

$$16x^{2}+x^{2}=\left(18-x\right)^{2}$$

Combine 16x^{2} and x^{2} to get 17x^{2}.

$$17x^{2}=\left(18-x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(18-x\right)^{2}.

$$17x^{2}=324-36x+x^{2}$$

Subtract 324 from both sides.

$$17x^{2}-324=-36x+x^{2}$$

Add 36x to both sides.

$$17x^{2}-324+36x=x^{2}$$

Subtract x^{2} from both sides.

$$17x^{2}-324+36x-x^{2}=0$$

Combine 17x^{2} and -x^{2} to get 16x^{2}.

$$16x^{2}-324+36x=0$$

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.

$$16x^{2}+36x-324=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, 36 for b, and -324 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-36±\sqrt{36^{2}-4\times 16\left(-324\right)}}{2\times 16}$$

Square 36.

$$x=\frac{-36±\sqrt{1296-4\times 16\left(-324\right)}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{-36±\sqrt{1296-64\left(-324\right)}}{2\times 16}$$

Multiply -64 times -324.

$$x=\frac{-36±\sqrt{1296+20736}}{2\times 16}$$

Add 1296 to 20736.

$$x=\frac{-36±\sqrt{22032}}{2\times 16}$$

Take the square root of 22032.

$$x=\frac{-36±36\sqrt{17}}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{-36±36\sqrt{17}}{32}$$

Now solve the equation x=\frac{-36±36\sqrt{17}}{32} when ± is plus. Add -36 to 36\sqrt{17}.

$$x=\frac{36\sqrt{17}-36}{32}$$

Divide -36+36\sqrt{17} by 32.

$$x=\frac{9\sqrt{17}-9}{8}$$

Now solve the equation x=\frac{-36±36\sqrt{17}}{32} when ± is minus. Subtract 36\sqrt{17} from -36.

$$x=\frac{-36\sqrt{17}-36}{32}$$

Divide -36-36\sqrt{17} by 32.

$$x=\frac{-9\sqrt{17}-9}{8}$$

The equation is now solved.

$$x=\frac{9\sqrt{17}-9}{8} x=\frac{-9\sqrt{17}-9}{8}$$

Steps for Completing the Square

Expand \left(4x\right)^{2}.

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

Calculate 4 to the power of 2 and get 16.

$$16x^{2}+x^{2}=\left(18-x\right)^{2}$$

Combine 16x^{2} and x^{2} to get 17x^{2}.

$$17x^{2}=\left(18-x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(18-x\right)^{2}.

$$17x^{2}=324-36x+x^{2}$$

Add 36x to both sides.

$$17x^{2}+36x=324+x^{2}$$

Subtract x^{2} from both sides.

$$17x^{2}+36x-x^{2}=324$$

Combine 17x^{2} and -x^{2} to get 16x^{2}.

$$16x^{2}+36x=324$$

Divide both sides by 16.

$$\frac{16x^{2}+36x}{16}=\frac{324}{16}$$

Dividing by 16 undoes the multiplication by 16.

$$x^{2}+\frac{36}{16}x=\frac{324}{16}$$

Reduce the fraction \frac{36}{16} to lowest terms by extracting and canceling out 4.

$$x^{2}+\frac{9}{4}x=\frac{324}{16}$$

Reduce the fraction \frac{324}{16} to lowest terms by extracting and canceling out 4.

$$x^{2}+\frac{9}{4}x=\frac{81}{4}$$

Divide \frac{9}{4}, the coefficient of the x term, by 2 to get \frac{9}{8}. Then add the square of \frac{9}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{9}{4}x+\left(\frac{9}{8}\right)^{2}=\frac{81}{4}+\left(\frac{9}{8}\right)^{2}$$

Square \frac{9}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{9}{4}x+\frac{81}{64}=\frac{81}{4}+\frac{81}{64}$$

Add \frac{81}{4} to \frac{81}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{9}{4}x+\frac{81}{64}=\frac{1377}{64}$$

Factor x^{2}+\frac{9}{4}x+\frac{81}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{9}{8}\right)^{2}=\frac{1377}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{9}{8}\right)^{2}}=\sqrt{\frac{1377}{64}}$$

Simplify.

$$x+\frac{9}{8}=\frac{9\sqrt{17}}{8} x+\frac{9}{8}=-\frac{9\sqrt{17}}{8}$$

Subtract \frac{9}{8} from both sides of the equation.

$$x=\frac{9\sqrt{17}-9}{8} x=\frac{-9\sqrt{17}-9}{8}$$