Solve 4+(b-1)^2=29

$$4 + ( b - 1 ) ^ { 2 } = 29$$

Solve for b

$$b=6$$

$$b=-4$$

Steps Using Factoring

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

$$4+b^{2}-2b+1=29$$

Add 4 and 1 to get 5.

$$5+b^{2}-2b=29$$

Subtract 29 from both sides.

$$5+b^{2}-2b-29=0$$

Subtract 29 from 5 to get -24.

$$-24+b^{2}-2b=0$$

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

$$b^{2}-2b-24=0$$

To solve the equation, factor b^{2}-2b-24 using formula b^{2}+\left(a+b\right)b+ab=\left(b+a\right)\left(b+b\right). To find a and b, set up a system to be solved.

$$a+b=-2 ab=-24$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -24.

$$1,-24 2,-12 3,-8 4,-6$$

Calculate the sum for each pair.

$$1-24=-23 2-12=-10 3-8=-5 4-6=-2$$

The solution is the pair that gives sum -2.

$$a=-6 b=4$$

Rewrite factored expression \left(b+a\right)\left(b+b\right) using the obtained values.

$$\left(b-6\right)\left(b+4\right)$$

To find equation solutions, solve b-6=0 and b+4=0.

$$b=6 b=-4$$

Steps Using Factoring By Grouping

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

$$4+b^{2}-2b+1=29$$

Add 4 and 1 to get 5.

$$5+b^{2}-2b=29$$

Subtract 29 from both sides.

$$5+b^{2}-2b-29=0$$

Subtract 29 from 5 to get -24.

$$-24+b^{2}-2b=0$$

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

$$b^{2}-2b-24=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as b^{2}+ab+bb-24. To find a and b, set up a system to be solved.

$$a+b=-2 ab=1\left(-24\right)=-24$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -24.

$$1,-24 2,-12 3,-8 4,-6$$

Calculate the sum for each pair.

$$1-24=-23 2-12=-10 3-8=-5 4-6=-2$$

The solution is the pair that gives sum -2.

$$a=-6 b=4$$

Rewrite b^{2}-2b-24 as \left(b^{2}-6b\right)+\left(4b-24\right).

$$\left(b^{2}-6b\right)+\left(4b-24\right)$$

Factor out b in the first and 4 in the second group.

$$b\left(b-6\right)+4\left(b-6\right)$$

Factor out common term b-6 by using distributive property.

$$\left(b-6\right)\left(b+4\right)$$

To find equation solutions, solve b-6=0 and b+4=0.

$$b=6 b=-4$$

Steps Using the Quadratic Formula

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

$$4+b^{2}-2b+1=29$$

Add 4 and 1 to get 5.

$$5+b^{2}-2b=29$$

Subtract 29 from both sides.

$$5+b^{2}-2b-29=0$$

Subtract 29 from 5 to get -24.

$$-24+b^{2}-2b=0$$

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.

$$b^{2}-2b-24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -2 for b, and -24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-24\right)}}{2}$$

Square -2.

$$b=\frac{-\left(-2\right)±\sqrt{4-4\left(-24\right)}}{2}$$

Multiply -4 times -24.

$$b=\frac{-\left(-2\right)±\sqrt{4+96}}{2}$$

Add 4 to 96.

$$b=\frac{-\left(-2\right)±\sqrt{100}}{2}$$

Take the square root of 100.

$$b=\frac{-\left(-2\right)±10}{2}$$

The opposite of -2 is 2.

$$b=\frac{2±10}{2}$$

Now solve the equation b=\frac{2±10}{2} when ± is plus. Add 2 to 10.

$$b=\frac{12}{2}$$

Divide 12 by 2.

$$b=6$$

Now solve the equation b=\frac{2±10}{2} when ± is minus. Subtract 10 from 2.

$$b=-\frac{8}{2}$$

Divide -8 by 2.

$$b=-4$$

The equation is now solved.

$$b=6 b=-4$$

Steps for Completing the Square

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

$$4+b^{2}-2b+1=29$$

Add 4 and 1 to get 5.

$$5+b^{2}-2b=29$$

Subtract 5 from both sides.

$$b^{2}-2b=29-5$$

Subtract 5 from 29 to get 24.

$$b^{2}-2b=24$$

Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$b^{2}-2b+1=24+1$$

Add 24 to 1.

$$b^{2}-2b+1=25$$

Factor b^{2}-2b+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(b-1\right)^{2}=25$$

Take the square root of both sides of the equation.

$$\sqrt{\left(b-1\right)^{2}}=\sqrt{25}$$

Simplify.

$$b-1=5 b-1=-5$$

Add 1 to both sides of the equation.

$$b=6 b=-4$$