Solve {[(2^{5})^{5}:4^{7}]x-13^{2}=(2x+2^{0})^2+(2^7)^5:4^15

$$\{ [ ( 2 ^ { 5 } ) ^ { 5 } : 4 ^ { 7 } ] x - 13 ^ { 2 } = ( 2 x + 2 ^ { 0 } ) ^ { 2 } + ( 2 ^ { 7 } ) ^ { 5 } : 4 ^ { 15 }$$

Solve for x

$$x = \frac{3 \sqrt{28991} + 511}{2} \approx 510.901155048$$

$$x=\frac{511-3\sqrt{28991}}{2}\approx 0.098844952$$

Steps Using the Quadratic Formula

To raise a power to another power, multiply the exponents. Multiply 5 and 5 to get 25.

$$\frac{2^{25}}{4^{7}}x-13^{2}=\left(2x+2^{0}\right)^{2}+\frac{\left(2^{7}\right)^{5}}{4^{15}}$$

To raise a power to another power, multiply the exponents. Multiply 7 and 5 to get 35.

$$\frac{2^{25}}{4^{7}}x-13^{2}=\left(2x+2^{0}\right)^{2}+\frac{2^{35}}{4^{15}}$$

Calculate 2 to the power of 25 and get 33554432.

$$\frac{33554432}{4^{7}}x-13^{2}=\left(2x+2^{0}\right)^{2}+\frac{2^{35}}{4^{15}}$$

Calculate 4 to the power of 7 and get 16384.

$$\frac{33554432}{16384}x-13^{2}=\left(2x+2^{0}\right)^{2}+\frac{2^{35}}{4^{15}}$$

Divide 33554432 by 16384 to get 2048.

$$2048x-13^{2}=\left(2x+2^{0}\right)^{2}+\frac{2^{35}}{4^{15}}$$

Calculate 13 to the power of 2 and get 169.

$$2048x-169=\left(2x+2^{0}\right)^{2}+\frac{2^{35}}{4^{15}}$$

Calculate 2 to the power of 0 and get 1.

$$2048x-169=\left(2x+1\right)^{2}+\frac{2^{35}}{4^{15}}$$

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

$$2048x-169=4x^{2}+4x+1+\frac{2^{35}}{4^{15}}$$

Calculate 2 to the power of 35 and get 34359738368.

$$2048x-169=4x^{2}+4x+1+\frac{34359738368}{4^{15}}$$

Calculate 4 to the power of 15 and get 1073741824.

$$2048x-169=4x^{2}+4x+1+\frac{34359738368}{1073741824}$$

Divide 34359738368 by 1073741824 to get 32.

$$2048x-169=4x^{2}+4x+1+32$$

Add 1 and 32 to get 33.

$$2048x-169=4x^{2}+4x+33$$

Subtract 4x^{2} from both sides.

$$2048x-169-4x^{2}=4x+33$$

Subtract 4x from both sides.

$$2048x-169-4x^{2}-4x=33$$

Combine 2048x and -4x to get 2044x.

$$2044x-169-4x^{2}=33$$

Subtract 33 from both sides.

$$2044x-169-4x^{2}-33=0$$

Subtract 33 from -169 to get -202.

$$2044x-202-4x^{2}=0$$

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.

$$-4x^{2}+2044x-202=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 2044 for b, and -202 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2044±\sqrt{2044^{2}-4\left(-4\right)\left(-202\right)}}{2\left(-4\right)}$$

Square 2044.

$$x=\frac{-2044±\sqrt{4177936-4\left(-4\right)\left(-202\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-2044±\sqrt{4177936+16\left(-202\right)}}{2\left(-4\right)}$$

Multiply 16 times -202.

$$x=\frac{-2044±\sqrt{4177936-3232}}{2\left(-4\right)}$$

Add 4177936 to -3232.

$$x=\frac{-2044±\sqrt{4174704}}{2\left(-4\right)}$$

Take the square root of 4174704.

$$x=\frac{-2044±12\sqrt{28991}}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-2044±12\sqrt{28991}}{-8}$$

Now solve the equation x=\frac{-2044±12\sqrt{28991}}{-8} when ± is plus. Add -2044 to 12\sqrt{28991}.

$$x=\frac{12\sqrt{28991}-2044}{-8}$$

Divide -2044+12\sqrt{28991} by -8.

$$x=\frac{511-3\sqrt{28991}}{2}$$

Now solve the equation x=\frac{-2044±12\sqrt{28991}}{-8} when ± is minus. Subtract 12\sqrt{28991} from -2044.

$$x=\frac{-12\sqrt{28991}-2044}{-8}$$

Divide -2044-12\sqrt{28991} by -8.

$$x=\frac{3\sqrt{28991}+511}{2}$$

The equation is now solved.

$$x=\frac{511-3\sqrt{28991}}{2} x=\frac{3\sqrt{28991}+511}{2}$$

Steps for Completing the Square

To raise a power to another power, multiply the exponents. Multiply 5 and 5 to get 25.

$$\frac{2^{25}}{4^{7}}x-13^{2}=\left(2x+2^{0}\right)^{2}+\frac{\left(2^{7}\right)^{5}}{4^{15}}$$

To raise a power to another power, multiply the exponents. Multiply 7 and 5 to get 35.

$$\frac{2^{25}}{4^{7}}x-13^{2}=\left(2x+2^{0}\right)^{2}+\frac{2^{35}}{4^{15}}$$

Calculate 2 to the power of 25 and get 33554432.

$$\frac{33554432}{4^{7}}x-13^{2}=\left(2x+2^{0}\right)^{2}+\frac{2^{35}}{4^{15}}$$

Calculate 4 to the power of 7 and get 16384.

$$\frac{33554432}{16384}x-13^{2}=\left(2x+2^{0}\right)^{2}+\frac{2^{35}}{4^{15}}$$

Divide 33554432 by 16384 to get 2048.

$$2048x-13^{2}=\left(2x+2^{0}\right)^{2}+\frac{2^{35}}{4^{15}}$$

Calculate 13 to the power of 2 and get 169.

$$2048x-169=\left(2x+2^{0}\right)^{2}+\frac{2^{35}}{4^{15}}$$

Calculate 2 to the power of 0 and get 1.

$$2048x-169=\left(2x+1\right)^{2}+\frac{2^{35}}{4^{15}}$$

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

$$2048x-169=4x^{2}+4x+1+\frac{2^{35}}{4^{15}}$$

Calculate 2 to the power of 35 and get 34359738368.

$$2048x-169=4x^{2}+4x+1+\frac{34359738368}{4^{15}}$$

Calculate 4 to the power of 15 and get 1073741824.

$$2048x-169=4x^{2}+4x+1+\frac{34359738368}{1073741824}$$

Divide 34359738368 by 1073741824 to get 32.

$$2048x-169=4x^{2}+4x+1+32$$

Add 1 and 32 to get 33.

$$2048x-169=4x^{2}+4x+33$$

Subtract 4x^{2} from both sides.

$$2048x-169-4x^{2}=4x+33$$

Subtract 4x from both sides.

$$2048x-169-4x^{2}-4x=33$$

Combine 2048x and -4x to get 2044x.

$$2044x-169-4x^{2}=33$$

Add 169 to both sides.

$$2044x-4x^{2}=33+169$$

Add 33 and 169 to get 202.

$$2044x-4x^{2}=202$$

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.

$$-4x^{2}+2044x=202$$

Divide both sides by -4.

$$\frac{-4x^{2}+2044x}{-4}=\frac{202}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\frac{2044}{-4}x=\frac{202}{-4}$$

Divide 2044 by -4.

$$x^{2}-511x=\frac{202}{-4}$$

Reduce the fraction \frac{202}{-4} to lowest terms by extracting and canceling out 2.

$$x^{2}-511x=-\frac{101}{2}$$

Divide -511, the coefficient of the x term, by 2 to get -\frac{511}{2}. Then add the square of -\frac{511}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-511x+\left(-\frac{511}{2}\right)^{2}=-\frac{101}{2}+\left(-\frac{511}{2}\right)^{2}$$

Square -\frac{511}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-511x+\frac{261121}{4}=-\frac{101}{2}+\frac{261121}{4}$$

Add -\frac{101}{2} to \frac{261121}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-511x+\frac{261121}{4}=\frac{260919}{4}$$

Factor x^{2}-511x+\frac{261121}{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}.

$$\left(x-\frac{511}{2}\right)^{2}=\frac{260919}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{511}{2}\right)^{2}}=\sqrt{\frac{260919}{4}}$$

Simplify.

$$x-\frac{511}{2}=\frac{3\sqrt{28991}}{2} x-\frac{511}{2}=-\frac{3\sqrt{28991}}{2}$$

Add \frac{511}{2} to both sides of the equation.

$$x=\frac{3\sqrt{28991}+511}{2} x=\frac{511-3\sqrt{28991}}{2}$$