How do you solve #x^ { 2} - 7x - 4= - 4#?

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Answer 1 · 2017-06-10T22:50:06

See a solution process below:

First, add #color(red)(4)# to each side of the equation to make the right side of the equation equal to #0#:

#x^2 - 7x - 4 + color(red)(4) = -4 + color(red)(4)#

#x^2 - 7x - 0 = 0#

#x^2 - 7x = 0#

Next, factor the left side of the equation as:

#(x * x) - (7 * x) = 0#

#x(x - 7) = 0#

Now, solve each term on the left side of the equation for #0#:

Solution 1)

#x = 0#

Solution 2)

#x - 7 = 0#

#x - 7 + color(red)(7) = 0 + color(red)(7)#

#x - 0 = 7#

#x = 7#

The solutions are: #x = 0# and #x = 7#