Solve 4(z-1)^2=z-1

$$4 ( z - 1 ) ^ { 2 } = z - 1$$

Solve for z

$$z = \frac{5}{4} = 1\frac{1}{4} = 1.25$$

$$z=1$$

Steps Using Factoring By Grouping

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

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

Use the distributive property to multiply 4 by z^{2}-2z+1.

$$4z^{2}-8z+4=z-1$$

Subtract z from both sides.

$$4z^{2}-8z+4-z=-1$$

Combine -8z and -z to get -9z.

$$4z^{2}-9z+4=-1$$

Add 1 to both sides.

$$4z^{2}-9z+4+1=0$$

Add 4 and 1 to get 5.

$$4z^{2}-9z+5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4z^{2}+az+bz+5. To find a and b, set up a system to be solved.

$$a+b=-9 ab=4\times 5=20$$

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 20.

$$-1,-20 -2,-10 -4,-5$$

Calculate the sum for each pair.

$$-1-20=-21 -2-10=-12 -4-5=-9$$

The solution is the pair that gives sum -9.

$$a=-5 b=-4$$

Rewrite 4z^{2}-9z+5 as \left(4z^{2}-5z\right)+\left(-4z+5\right).

$$\left(4z^{2}-5z\right)+\left(-4z+5\right)$$

Factor out z in the first and -1 in the second group.

$$z\left(4z-5\right)-\left(4z-5\right)$$

Factor out common term 4z-5 by using distributive property.

$$\left(4z-5\right)\left(z-1\right)$$

To find equation solutions, solve 4z-5=0 and z-1=0.

$$z=\frac{5}{4} z=1$$

Steps Using the Quadratic Formula

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

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

Use the distributive property to multiply 4 by z^{2}-2z+1.

$$4z^{2}-8z+4=z-1$$

Subtract z from both sides.

$$4z^{2}-8z+4-z=-1$$

Combine -8z and -z to get -9z.

$$4z^{2}-9z+4=-1$$

Add 1 to both sides.

$$4z^{2}-9z+4+1=0$$

Add 4 and 1 to get 5.

$$4z^{2}-9z+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -9 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$z=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 4\times 5}}{2\times 4}$$

Square -9.

$$z=\frac{-\left(-9\right)±\sqrt{81-4\times 4\times 5}}{2\times 4}$$

Multiply -4 times 4.

$$z=\frac{-\left(-9\right)±\sqrt{81-16\times 5}}{2\times 4}$$

Multiply -16 times 5.

$$z=\frac{-\left(-9\right)±\sqrt{81-80}}{2\times 4}$$

Add 81 to -80.

$$z=\frac{-\left(-9\right)±\sqrt{1}}{2\times 4}$$

Take the square root of 1.

$$z=\frac{-\left(-9\right)±1}{2\times 4}$$

The opposite of -9 is 9.

$$z=\frac{9±1}{2\times 4}$$

Multiply 2 times 4.

$$z=\frac{9±1}{8}$$

Now solve the equation z=\frac{9±1}{8} when ± is plus. Add 9 to 1.

$$z=\frac{10}{8}$$

Reduce the fraction \frac{10}{8} to lowest terms by extracting and canceling out 2.

$$z=\frac{5}{4}$$

Now solve the equation z=\frac{9±1}{8} when ± is minus. Subtract 1 from 9.

$$z=\frac{8}{8}$$

Divide 8 by 8.

$$z=1$$

The equation is now solved.

$$z=\frac{5}{4} z=1$$

Steps for Completing the Square

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

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

Use the distributive property to multiply 4 by z^{2}-2z+1.

$$4z^{2}-8z+4=z-1$$

Subtract z from both sides.

$$4z^{2}-8z+4-z=-1$$

Combine -8z and -z to get -9z.

$$4z^{2}-9z+4=-1$$

Subtract 4 from both sides.

$$4z^{2}-9z=-1-4$$

Subtract 4 from -1 to get -5.

$$4z^{2}-9z=-5$$

Divide both sides by 4.

$$\frac{4z^{2}-9z}{4}=-\frac{5}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$z^{2}-\frac{9}{4}z=-\frac{5}{4}$$

Divide -\frac{9}{4}, the coefficient of the x term, by 2 to get -\frac{9}{8}. Then add the square of -\frac{9}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$z^{2}-\frac{9}{4}z+\left(-\frac{9}{8}\right)^{2}=-\frac{5}{4}+\left(-\frac{9}{8}\right)^{2}$$

Square -\frac{9}{8} by squaring both the numerator and the denominator of the fraction.

$$z^{2}-\frac{9}{4}z+\frac{81}{64}=-\frac{5}{4}+\frac{81}{64}$$

Add -\frac{5}{4} to \frac{81}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$z^{2}-\frac{9}{4}z+\frac{81}{64}=\frac{1}{64}$$

Factor z^{2}-\frac{9}{4}z+\frac{81}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(z-\frac{9}{8}\right)^{2}=\frac{1}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(z-\frac{9}{8}\right)^{2}}=\sqrt{\frac{1}{64}}$$

Simplify.

$$z-\frac{9}{8}=\frac{1}{8} z-\frac{9}{8}=-\frac{1}{8}$$

Add \frac{9}{8} to both sides of the equation.

$$z=\frac{5}{4} z=1$$