# How do you solve #49b^2 + 84b + 32 = 0 # by completing the square?

##### 3 Answers

#### Explanation:

The first step is to find the number that squares to give 49, which is 7. Then, you need to find a number such that

After that, it's a case of square rooting both sides, then rearranging to get the answer. Since both

#### Explanation:

The difference of squares identity can be written:

#A^2-B^2=(A-B)(A+B)#

We will use this with

Note that

So we find:

#0 = 49b^2+84b+32#

#color(white)(0) = (7b)^2+2(7b)(6)+(6)^2-4#

#color(white)(0) = (7b)^2+2(7b)(6)+(6)^2-4#

#color(white)(0) = (7b+6)^2-2^2#

#color(white)(0) = ((7b+6)-2)((7b+6)+2)#

#color(white)(0) = (7b+4)(7b+8)#

Hence:

#b = -4/7" "# or#" "b = -8/7#

#### Explanation:

This is the method I was taught to complete the square on a general quadratic:

# x^2+bx+c #

- Step 1: Factor out (or divide) the coefficient of
#x^2# so that that coefficient is#1# . - Step 2:Use the knowledge of a perfect square
#(x+alpha)^2=x^2+2alphax+alpha^2# , Here we have#2alpha=b=>alpha=1/2b# and subtract#alpha^2=(1/2b)^2# - Step 3: Solve the equation

So for this particular problem we have

**Step 1:** Divide by

# 49b^2 + 84b + 32 = 0 #

# :. b^2 + 84/49b + 32/49 = 0 #

# :. b^2 + 12/7b + 32/49 = 0 #

**Step 2:** Form a perfect square using

# :. (b + 1/2*12/7)^2 - (1/2*12/7)^2 + 32/49=0 #

# :. (b + 6/7)^2 - (6/7)^2 + 32/49=0 #

# :. (b + 6/7)^2 - 36/49 + 32/49=0 #

# :. (b + 6/7)^2 - 4/49=0 #

**Step 3:**: If we are solving an equation then solve it

# :. (b + 6/7)^2 = 4/49 #

# :. b + 6/7 = +-sqrt(4/49) #

# :. b + 6/7 = +-2/7 #

# :. b = - 6/7 +-2/7 #

Leading to the two solutions:

# b = - 6/7 -2/7 = -8/7#

# b = - 6/7 +2/7 = -4/7#