Solve a-1/a^2-2a+1=3/1

$$\frac { a - 1 } { a ^ { 2 } - 2 a + 1 } = \frac { 3 } { 1 }$$

Solve for a

$$a = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

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 \left(a-1\right)^{2}.

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

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

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

Use the distributive property to multiply a^{2}-2a+1 by 3.

$$a-1=3a^{2}-6a+3$$

Subtract 3a^{2} from both sides.

$$a-1-3a^{2}=-6a+3$$

Add 6a to both sides.

$$a-1-3a^{2}+6a=3$$

Combine a and 6a to get 7a.

$$7a-1-3a^{2}=3$$

Subtract 3 from both sides.

$$7a-1-3a^{2}-3=0$$

Subtract 3 from -1 to get -4.

$$7a-4-3a^{2}=0$$

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

$$-3a^{2}+7a-4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3a^{2}+aa+ba-4. To find a and b, set up a system to be solved.

$$a+b=7 ab=-3\left(-4\right)=12$$

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 12.

$$1,12 2,6 3,4$$

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 7.

$$a=4 b=3$$

Rewrite -3a^{2}+7a-4 as \left(-3a^{2}+4a\right)+\left(3a-4\right).

$$\left(-3a^{2}+4a\right)+\left(3a-4\right)$$

Factor out -a in -3a^{2}+4a.

$$-a\left(3a-4\right)+3a-4$$

Factor out common term 3a-4 by using distributive property.

$$\left(3a-4\right)\left(-a+1\right)$$

To find equation solutions, solve 3a-4=0 and -a+1=0.

$$a=\frac{4}{3} a=1$$

Variable a cannot be equal to 1.

$$a=\frac{4}{3}$$

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 \left(a-1\right)^{2}.

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

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

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

Use the distributive property to multiply a^{2}-2a+1 by 3.

$$a-1=3a^{2}-6a+3$$

Subtract 3a^{2} from both sides.

$$a-1-3a^{2}=-6a+3$$

Add 6a to both sides.

$$a-1-3a^{2}+6a=3$$

Combine a and 6a to get 7a.

$$7a-1-3a^{2}=3$$

Subtract 3 from both sides.

$$7a-1-3a^{2}-3=0$$

Subtract 3 from -1 to get -4.

$$7a-4-3a^{2}=0$$

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.

$$-3a^{2}+7a-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 7 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-7±\sqrt{7^{2}-4\left(-3\right)\left(-4\right)}}{2\left(-3\right)}$$

Square 7.

$$a=\frac{-7±\sqrt{49-4\left(-3\right)\left(-4\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$a=\frac{-7±\sqrt{49+12\left(-4\right)}}{2\left(-3\right)}$$

Multiply 12 times -4.

$$a=\frac{-7±\sqrt{49-48}}{2\left(-3\right)}$$

Add 49 to -48.

$$a=\frac{-7±\sqrt{1}}{2\left(-3\right)}$$

Take the square root of 1.

$$a=\frac{-7±1}{2\left(-3\right)}$$

Multiply 2 times -3.

$$a=\frac{-7±1}{-6}$$

Now solve the equation a=\frac{-7±1}{-6} when ± is plus. Add -7 to 1.

$$a=-\frac{6}{-6}$$

Divide -6 by -6.

$$a=1$$

Now solve the equation a=\frac{-7±1}{-6} when ± is minus. Subtract 1 from -7.

$$a=-\frac{8}{-6}$$

Reduce the fraction \frac{-8}{-6} to lowest terms by extracting and canceling out 2.

$$a=\frac{4}{3}$$

The equation is now solved.

$$a=1 a=\frac{4}{3}$$

Variable a cannot be equal to 1.

$$a=\frac{4}{3}$$

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 \left(a-1\right)^{2}.

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

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

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

Use the distributive property to multiply a^{2}-2a+1 by 3.

$$a-1=3a^{2}-6a+3$$

Subtract 3a^{2} from both sides.

$$a-1-3a^{2}=-6a+3$$

Add 6a to both sides.

$$a-1-3a^{2}+6a=3$$

Combine a and 6a to get 7a.

$$7a-1-3a^{2}=3$$

Add 1 to both sides.

$$7a-3a^{2}=3+1$$

Add 3 and 1 to get 4.

$$7a-3a^{2}=4$$

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.

$$-3a^{2}+7a=4$$

Divide both sides by -3.

$$\frac{-3a^{2}+7a}{-3}=\frac{4}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$a^{2}+\frac{7}{-3}a=\frac{4}{-3}$$

Divide 7 by -3.

$$a^{2}-\frac{7}{3}a=\frac{4}{-3}$$

Divide 4 by -3.

$$a^{2}-\frac{7}{3}a=-\frac{4}{3}$$

Divide -\frac{7}{3}, the coefficient of the x term, by 2 to get -\frac{7}{6}. Then add the square of -\frac{7}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}-\frac{7}{3}a+\left(-\frac{7}{6}\right)^{2}=-\frac{4}{3}+\left(-\frac{7}{6}\right)^{2}$$

Square -\frac{7}{6} by squaring both the numerator and the denominator of the fraction.

$$a^{2}-\frac{7}{3}a+\frac{49}{36}=-\frac{4}{3}+\frac{49}{36}$$

Add -\frac{4}{3} to \frac{49}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a^{2}-\frac{7}{3}a+\frac{49}{36}=\frac{1}{36}$$

Factor a^{2}-\frac{7}{3}a+\frac{49}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(a-\frac{7}{6}\right)^{2}=\frac{1}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-\frac{7}{6}\right)^{2}}=\sqrt{\frac{1}{36}}$$

Simplify.

$$a-\frac{7}{6}=\frac{1}{6} a-\frac{7}{6}=-\frac{1}{6}$$

Add \frac{7}{6} to both sides of the equation.

$$a=\frac{4}{3} a=1$$

Variable a cannot be equal to 1.

$$a=\frac{4}{3}$$