Solve (4a-1/a+1-a+1)=a^2-8a+16/a+1

$$( \frac { 4 a - 1 } { a + 1 } - a + 1 ) = \frac { a ^ { 2 } - 8 a + 16 } { a + 1 }$$

Solve for a

$$a=2$$

$$a=4$$

Steps Using Factoring By Grouping

Variable a cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by a+1.

$$4a-1-a\left(a+1\right)+a+1=a^{2}-8a+16$$

Use the distributive property to multiply -a by a+1.

$$4a-1-a^{2}-a+a+1=a^{2}-8a+16$$

Combine 4a and -a to get 3a.

$$3a-1-a^{2}+a+1=a^{2}-8a+16$$

Combine 3a and a to get 4a.

$$4a-1-a^{2}+1=a^{2}-8a+16$$

Add -1 and 1 to get 0.

$$4a-a^{2}=a^{2}-8a+16$$

Subtract a^{2} from both sides.

$$4a-a^{2}-a^{2}=-8a+16$$

Combine -a^{2} and -a^{2} to get -2a^{2}.

$$4a-2a^{2}=-8a+16$$

Add 8a to both sides.

$$4a-2a^{2}+8a=16$$

Combine 4a and 8a to get 12a.

$$12a-2a^{2}=16$$

Subtract 16 from both sides.

$$12a-2a^{2}-16=0$$

Divide both sides by 2.

$$6a-a^{2}-8=0$$

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

$$-a^{2}+6a-8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -a^{2}+aa+ba-8. To find a and b, set up a system to be solved.

$$a+b=6 ab=-\left(-8\right)=8$$

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 8.

$$1,8 2,4$$

Calculate the sum for each pair.

$$1+8=9 2+4=6$$

The solution is the pair that gives sum 6.

$$a=4 b=2$$

Rewrite -a^{2}+6a-8 as \left(-a^{2}+4a\right)+\left(2a-8\right).

$$\left(-a^{2}+4a\right)+\left(2a-8\right)$$

Factor out -a in the first and 2 in the second group.

$$-a\left(a-4\right)+2\left(a-4\right)$$

Factor out common term a-4 by using distributive property.

$$\left(a-4\right)\left(-a+2\right)$$

To find equation solutions, solve a-4=0 and -a+2=0.

$$a=4 a=2$$

Steps Using the Quadratic Formula

Variable a cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by a+1.

$$4a-1-a\left(a+1\right)+a+1=a^{2}-8a+16$$

Use the distributive property to multiply -a by a+1.

$$4a-1-a^{2}-a+a+1=a^{2}-8a+16$$

Combine 4a and -a to get 3a.

$$3a-1-a^{2}+a+1=a^{2}-8a+16$$

Combine 3a and a to get 4a.

$$4a-1-a^{2}+1=a^{2}-8a+16$$

Add -1 and 1 to get 0.

$$4a-a^{2}=a^{2}-8a+16$$

Subtract a^{2} from both sides.

$$4a-a^{2}-a^{2}=-8a+16$$

Combine -a^{2} and -a^{2} to get -2a^{2}.

$$4a-2a^{2}=-8a+16$$

Add 8a to both sides.

$$4a-2a^{2}+8a=16$$

Combine 4a and 8a to get 12a.

$$12a-2a^{2}=16$$

Subtract 16 from both sides.

$$12a-2a^{2}-16=0$$

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.

$$-2a^{2}+12a-16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 12 for b, and -16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-12±\sqrt{12^{2}-4\left(-2\right)\left(-16\right)}}{2\left(-2\right)}$$

Square 12.

$$a=\frac{-12±\sqrt{144-4\left(-2\right)\left(-16\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$a=\frac{-12±\sqrt{144+8\left(-16\right)}}{2\left(-2\right)}$$

Multiply 8 times -16.

$$a=\frac{-12±\sqrt{144-128}}{2\left(-2\right)}$$

Add 144 to -128.

$$a=\frac{-12±\sqrt{16}}{2\left(-2\right)}$$

Take the square root of 16.

$$a=\frac{-12±4}{2\left(-2\right)}$$

Multiply 2 times -2.

$$a=\frac{-12±4}{-4}$$

Now solve the equation a=\frac{-12±4}{-4} when ± is plus. Add -12 to 4.

$$a=-\frac{8}{-4}$$

Divide -8 by -4.

$$a=2$$

Now solve the equation a=\frac{-12±4}{-4} when ± is minus. Subtract 4 from -12.

$$a=-\frac{16}{-4}$$

Divide -16 by -4.

$$a=4$$

The equation is now solved.

$$a=2 a=4$$

Steps for Completing the Square

Variable a cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by a+1.

$$4a-1-a\left(a+1\right)+a+1=a^{2}-8a+16$$

Use the distributive property to multiply -a by a+1.

$$4a-1-a^{2}-a+a+1=a^{2}-8a+16$$

Combine 4a and -a to get 3a.

$$3a-1-a^{2}+a+1=a^{2}-8a+16$$

Combine 3a and a to get 4a.

$$4a-1-a^{2}+1=a^{2}-8a+16$$

Add -1 and 1 to get 0.

$$4a-a^{2}=a^{2}-8a+16$$

Subtract a^{2} from both sides.

$$4a-a^{2}-a^{2}=-8a+16$$

Combine -a^{2} and -a^{2} to get -2a^{2}.

$$4a-2a^{2}=-8a+16$$

Add 8a to both sides.

$$4a-2a^{2}+8a=16$$

Combine 4a and 8a to get 12a.

$$12a-2a^{2}=16$$

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.

$$-2a^{2}+12a=16$$

Divide both sides by -2.

$$\frac{-2a^{2}+12a}{-2}=\frac{16}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$a^{2}+\frac{12}{-2}a=\frac{16}{-2}$$

Divide 12 by -2.

$$a^{2}-6a=\frac{16}{-2}$$

Divide 16 by -2.

$$a^{2}-6a=-8$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}-6a+\left(-3\right)^{2}=-8+\left(-3\right)^{2}$$

Square -3.

$$a^{2}-6a+9=-8+9$$

Add -8 to 9.

$$a^{2}-6a+9=1$$

Factor a^{2}-6a+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(a-3\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-3\right)^{2}}=\sqrt{1}$$

Simplify.

$$a-3=1 a-3=-1$$

Add 3 to both sides of the equation.

$$a=4 a=2$$