Solve -(x-4)^2+1=0

$$- ( x - 4 ) ^ { 2 } + 1 = 0$$

Solve for x

$$x=5$$

$$x=3$$

Steps Using Factoring By Grouping

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

$$-\left(x^{2}-8x+16\right)+1=0$$

To find the opposite of x^{2}-8x+16, find the opposite of each term.

$$-x^{2}+8x-16+1=0$$

Add -16 and 1 to get -15.

$$-x^{2}+8x-15=0$$

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

$$a+b=8 ab=-\left(-15\right)=15$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 15.

$$1,15 3,5$$

Calculate the sum for each pair.

$$1+15=16 3+5=8$$

The solution is the pair that gives sum 8.

$$a=5 b=3$$

Rewrite -x^{2}+8x-15 as \left(-x^{2}+5x\right)+\left(3x-15\right).

$$\left(-x^{2}+5x\right)+\left(3x-15\right)$$

Factor out -x in the first and 3 in the second group.

$$-x\left(x-5\right)+3\left(x-5\right)$$

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

$$\left(x-5\right)\left(-x+3\right)$$

To find equation solutions, solve x-5=0 and -x+3=0.

$$x=5 x=3$$

Steps Using the Quadratic Formula

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

$$-\left(x^{2}-8x+16\right)+1=0$$

To find the opposite of x^{2}-8x+16, find the opposite of each term.

$$-x^{2}+8x-16+1=0$$

Add -16 and 1 to get -15.

$$-x^{2}+8x-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 8 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-1\right)\left(-15\right)}}{2\left(-1\right)}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\left(-1\right)\left(-15\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-8±\sqrt{64+4\left(-15\right)}}{2\left(-1\right)}$$

Multiply 4 times -15.

$$x=\frac{-8±\sqrt{64-60}}{2\left(-1\right)}$$

Add 64 to -60.

$$x=\frac{-8±\sqrt{4}}{2\left(-1\right)}$$

Take the square root of 4.

$$x=\frac{-8±2}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-8±2}{-2}$$

Now solve the equation x=\frac{-8±2}{-2} when ± is plus. Add -8 to 2.

$$x=-\frac{6}{-2}$$

Divide -6 by -2.

$$x=3$$

Now solve the equation x=\frac{-8±2}{-2} when ± is minus. Subtract 2 from -8.

$$x=-\frac{10}{-2}$$

Divide -10 by -2.

$$x=5$$

The equation is now solved.

$$x=3 x=5$$

Steps for Completing the Square

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

$$-\left(x^{2}-8x+16\right)+1=0$$

To find the opposite of x^{2}-8x+16, find the opposite of each term.

$$-x^{2}+8x-16+1=0$$

Add -16 and 1 to get -15.

$$-x^{2}+8x-15=0$$

Add 15 to both sides. Anything plus zero gives itself.

$$-x^{2}+8x=15$$

Divide both sides by -1.

$$\frac{-x^{2}+8x}{-1}=\frac{15}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{8}{-1}x=\frac{15}{-1}$$

Divide 8 by -1.

$$x^{2}-8x=\frac{15}{-1}$$

Divide 15 by -1.

$$x^{2}-8x=-15$$

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

$$x^{2}-8x+\left(-4\right)^{2}=-15+\left(-4\right)^{2}$$

Square -4.

$$x^{2}-8x+16=-15+16$$

Add -15 to 16.

$$x^{2}-8x+16=1$$

Factor x^{2}-8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-4\right)^{2}=1$$

Take the square root of both sides of the equation.

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

Simplify.

$$x-4=1 x-4=-1$$

Add 4 to both sides of the equation.

$$x=5 x=3$$