How do you solve #4x ^ { 2} - 24x - 28= 0#?

1 Answer
AA_123
Nov 15, 2016

#x=-1# or #x=7#

Explanation:

We can begin by dividing both sides by four

#(4x^2-24x-28)/4=0/4#

#=x^2color(red)-color(red)6xcolor (blue)-color (blue)7=0#

Then we factor the polynomial to find what values of #x# make the expression equal to zero. To factor we ask our selves: What two numbers if we take their product we will get #color (blue)-color(blue)7# and if we take their sum we will get #color (red)-color (red)6#?

The two numbers are #1# and #-7#. So

#(x+1)(x-7)=0#

in order of the left side to be equal to zero one of the two expressions inside the brackets must be equal to zero, because if one of them was zero then the whole expression will be equal to zero since we are multiplying by zero. So

The first answer:-
1) #x+1=0#
"#x=-1#"

The second answer:-
2) #x-7=0#
"#x=7#"

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