To find the solution set of a factored quadratic equation, you should set each one of the factors equal to zero and solve for [tex]x[/tex].
Lets take our first quadratic equation (x+1/2)(x-5)=0 and apply this. First we are our factors equal to zero [tex]x+ \frac{1}{2} =0[/tex] and [tex]x-5=0[/tex]; next we are going to sole for [tex]x[/tex] in each factor to find the solution set:
[tex]x=- \frac{1}{2} [/tex] and [tex]x=5[/tex], so the solution set for the quadratic equation (x+1/2)(x-5)=0 is [tex](- \frac{1}{2},5)[/tex]
Lets do the same for our next one (x-5)(2x-1)=0
[tex]x-5=0[/tex]
[tex]x=5[/tex]
and
[tex]2x-1=0[/tex]
[tex]2x=1[/tex]
[tex]x= \frac{1}{2} [/tex]
So, the solution set for the quadratic equation (x-5)(2x-1)=0 is [tex](5, \frac{1}{2}) [/tex]
Next one (x+5)(2x-1)=0
[tex]x+5=0[/tex]
[tex]x=-5[/tex]
and
[tex]2x-1=0[/tex]
[tex]2x=1[/tex]
[tex]x= \frac{1}{2} [/tex]
So, the solution set for the quadratic equation (x+5)(2x-1)=0 is [tex](-5, \frac{1}{2} )[/tex]
Next one (-2x+1)(-x+5)=0
[tex]-2x+1=0[/tex]
[tex]-2x=-1[/tex]
[tex]x= \frac{-1}{-2} [/tex]
[tex]x= \frac{1}{2} [/tex]
and
[tex]-x+5=0[/tex]
[tex]-x=-5[/tex]
[tex]x=5[/tex]
So, the solution set for the quadratic equation (-2x+1)(-x+5)=0 is [tex]( \frac{1}{2} ,5)[/tex]: therefore, we are going the select this one.
Next one (x+1/2)(x+5)=0
[tex]x+ \frac{1}{2} =0[/tex]
[tex]x=- \frac{1}{2} [/tex]
and
[tex]x+5=0[/tex]
[tex]x=-5[/tex]
So, the solution set for the quadratic equation (x+1/2)(x+5)=0 is [tex](- \frac{1}{2} ,-5)[/tex]
Finally, our last one (-2x+1)(x-5)=0
[tex]-2x+1=0[/tex]
[tex]-2x=-1[/tex]
[tex]x= \frac{-1}{-2} [/tex]
[tex]x= \frac{1}{2} [/tex]
and
[tex]x-5=0[/tex]
[tex]x=5[/tex]
So, the solution set for the quadratic equation (-2x+1)(x-5)=0 is [tex]( \frac{1}{2} ,5)[/tex]; this is also a correct answer, make sure to select this one too.
We can conclude that both (-2x+1)(-x+5)=0 and (-2x+1)(x-5)=0 quadratic equation have [tex]( \frac{1}{2} ,5)[/tex] as solution set.
