Solve x^2+(8-2x)^2=29

$$x ^ { 2 } + ( 8 - 2 x ) ^ { 2 } = 29$$

Solve for x

$$x=5$$

$$x = \frac{7}{5} = 1\frac{2}{5} = 1.4$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(8-2x\right)^{2}.

$$x^{2}+64-32x+4x^{2}=29$$

Combine x^{2} and 4x^{2} to get 5x^{2}.

$$5x^{2}+64-32x=29$$

Subtract 29 from both sides.

$$5x^{2}+64-32x-29=0$$

Subtract 29 from 64 to get 35.

$$5x^{2}+35-32x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$5x^{2}-32x+35=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx+35. To find a and b, set up a system to be solved.

$$a+b=-32 ab=5\times 35=175$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 175.

$$-1,-175 -5,-35 -7,-25$$

Calculate the sum for each pair.

$$-1-175=-176 -5-35=-40 -7-25=-32$$

The solution is the pair that gives sum -32.

$$a=-25 b=-7$$

Rewrite 5x^{2}-32x+35 as \left(5x^{2}-25x\right)+\left(-7x+35\right).

$$\left(5x^{2}-25x\right)+\left(-7x+35\right)$$

Factor out 5x in the first and -7 in the second group.

$$5x\left(x-5\right)-7\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(5x-7\right)$$

To find equation solutions, solve x-5=0 and 5x-7=0.

$$x=5 x=\frac{7}{5}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(8-2x\right)^{2}.

$$x^{2}+64-32x+4x^{2}=29$$

Combine x^{2} and 4x^{2} to get 5x^{2}.

$$5x^{2}+64-32x=29$$

Subtract 29 from both sides.

$$5x^{2}+64-32x-29=0$$

Subtract 29 from 64 to get 35.

$$5x^{2}+35-32x=0$$

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.

$$5x^{2}-32x+35=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -32 for b, and 35 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-32\right)±\sqrt{\left(-32\right)^{2}-4\times 5\times 35}}{2\times 5}$$

Square -32.

$$x=\frac{-\left(-32\right)±\sqrt{1024-4\times 5\times 35}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-32\right)±\sqrt{1024-20\times 35}}{2\times 5}$$

Multiply -20 times 35.

$$x=\frac{-\left(-32\right)±\sqrt{1024-700}}{2\times 5}$$

Add 1024 to -700.

$$x=\frac{-\left(-32\right)±\sqrt{324}}{2\times 5}$$

Take the square root of 324.

$$x=\frac{-\left(-32\right)±18}{2\times 5}$$

The opposite of -32 is 32.

$$x=\frac{32±18}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{32±18}{10}$$

Now solve the equation x=\frac{32±18}{10} when ± is plus. Add 32 to 18.

$$x=\frac{50}{10}$$

Divide 50 by 10.

$$x=5$$

Now solve the equation x=\frac{32±18}{10} when ± is minus. Subtract 18 from 32.

$$x=\frac{14}{10}$$

Reduce the fraction \frac{14}{10} to lowest terms by extracting and canceling out 2.

$$x=\frac{7}{5}$$

The equation is now solved.

$$x=5 x=\frac{7}{5}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(8-2x\right)^{2}.

$$x^{2}+64-32x+4x^{2}=29$$

Combine x^{2} and 4x^{2} to get 5x^{2}.

$$5x^{2}+64-32x=29$$

Subtract 64 from both sides.

$$5x^{2}-32x=29-64$$

Subtract 64 from 29 to get -35.

$$5x^{2}-32x=-35$$

Divide both sides by 5.

$$\frac{5x^{2}-32x}{5}=-\frac{35}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{32}{5}x=-\frac{35}{5}$$

Divide -35 by 5.

$$x^{2}-\frac{32}{5}x=-7$$

Divide -\frac{32}{5}, the coefficient of the x term, by 2 to get -\frac{16}{5}. Then add the square of -\frac{16}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{32}{5}x+\left(-\frac{16}{5}\right)^{2}=-7+\left(-\frac{16}{5}\right)^{2}$$

Square -\frac{16}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{32}{5}x+\frac{256}{25}=-7+\frac{256}{25}$$

Add -7 to \frac{256}{25}.

$$x^{2}-\frac{32}{5}x+\frac{256}{25}=\frac{81}{25}$$

Factor x^{2}-\frac{32}{5}x+\frac{256}{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}.

$$\left(x-\frac{16}{5}\right)^{2}=\frac{81}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{16}{5}\right)^{2}}=\sqrt{\frac{81}{25}}$$

Simplify.

$$x-\frac{16}{5}=\frac{9}{5} x-\frac{16}{5}=-\frac{9}{5}$$

Add \frac{16}{5} to both sides of the equation.

$$x=5 x=\frac{7}{5}$$