Solve (4-4x)^2+x^2=1

$$( 4 - 4 x ) ^ { 2 } + x ^ { 2 } = 1$$

Solve for x

$$x=1$$

$$x=\frac{15}{17}\approx 0.882352941$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-4x\right)^{2}.

$$16-32x+16x^{2}+x^{2}=1$$

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

$$16-32x+17x^{2}=1$$

Subtract 1 from both sides.

$$16-32x+17x^{2}-1=0$$

Subtract 1 from 16 to get 15.

$$15-32x+17x^{2}=0$$

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

$$17x^{2}-32x+15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 17x^{2}+ax+bx+15. To find a and b, set up a system to be solved.

$$a+b=-32 ab=17\times 15=255$$

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 255.

$$-1,-255 -3,-85 -5,-51 -15,-17$$

Calculate the sum for each pair.

$$-1-255=-256 -3-85=-88 -5-51=-56 -15-17=-32$$

The solution is the pair that gives sum -32.

$$a=-17 b=-15$$

Rewrite 17x^{2}-32x+15 as \left(17x^{2}-17x\right)+\left(-15x+15\right).

$$\left(17x^{2}-17x\right)+\left(-15x+15\right)$$

Factor out 17x in the first and -15 in the second group.

$$17x\left(x-1\right)-15\left(x-1\right)$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(17x-15\right)$$

To find equation solutions, solve x-1=0 and 17x-15=0.

$$x=1 x=\frac{15}{17}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-4x\right)^{2}.

$$16-32x+16x^{2}+x^{2}=1$$

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

$$16-32x+17x^{2}=1$$

Subtract 1 from both sides.

$$16-32x+17x^{2}-1=0$$

Subtract 1 from 16 to get 15.

$$15-32x+17x^{2}=0$$

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.

$$17x^{2}-32x+15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 17 for a, -32 for b, and 15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-32\right)±\sqrt{\left(-32\right)^{2}-4\times 17\times 15}}{2\times 17}$$

Square -32.

$$x=\frac{-\left(-32\right)±\sqrt{1024-4\times 17\times 15}}{2\times 17}$$

Multiply -4 times 17.

$$x=\frac{-\left(-32\right)±\sqrt{1024-68\times 15}}{2\times 17}$$

Multiply -68 times 15.

$$x=\frac{-\left(-32\right)±\sqrt{1024-1020}}{2\times 17}$$

Add 1024 to -1020.

$$x=\frac{-\left(-32\right)±\sqrt{4}}{2\times 17}$$

Take the square root of 4.

$$x=\frac{-\left(-32\right)±2}{2\times 17}$$

The opposite of -32 is 32.

$$x=\frac{32±2}{2\times 17}$$

Multiply 2 times 17.

$$x=\frac{32±2}{34}$$

Now solve the equation x=\frac{32±2}{34} when ± is plus. Add 32 to 2.

$$x=\frac{34}{34}$$

Divide 34 by 34.

$$x=1$$

Now solve the equation x=\frac{32±2}{34} when ± is minus. Subtract 2 from 32.

$$x=\frac{30}{34}$$

Reduce the fraction \frac{30}{34} to lowest terms by extracting and canceling out 2.

$$x=\frac{15}{17}$$

The equation is now solved.

$$x=1 x=\frac{15}{17}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-4x\right)^{2}.

$$16-32x+16x^{2}+x^{2}=1$$

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

$$16-32x+17x^{2}=1$$

Subtract 16 from both sides.

$$-32x+17x^{2}=1-16$$

Subtract 16 from 1 to get -15.

$$-32x+17x^{2}=-15$$

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.

$$17x^{2}-32x=-15$$

Divide both sides by 17.

$$\frac{17x^{2}-32x}{17}=-\frac{15}{17}$$

Dividing by 17 undoes the multiplication by 17.

$$x^{2}-\frac{32}{17}x=-\frac{15}{17}$$

Divide -\frac{32}{17}, the coefficient of the x term, by 2 to get -\frac{16}{17}. Then add the square of -\frac{16}{17} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{32}{17}x+\left(-\frac{16}{17}\right)^{2}=-\frac{15}{17}+\left(-\frac{16}{17}\right)^{2}$$

Square -\frac{16}{17} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{32}{17}x+\frac{256}{289}=-\frac{15}{17}+\frac{256}{289}$$

Add -\frac{15}{17} to \frac{256}{289} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{32}{17}x+\frac{256}{289}=\frac{1}{289}$$

Factor x^{2}-\frac{32}{17}x+\frac{256}{289}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{16}{17}\right)^{2}=\frac{1}{289}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{16}{17}\right)^{2}}=\sqrt{\frac{1}{289}}$$

Simplify.

$$x-\frac{16}{17}=\frac{1}{17} x-\frac{16}{17}=-\frac{1}{17}$$

Add \frac{16}{17} to both sides of the equation.

$$x=1 x=\frac{15}{17}$$