Respuesta :
(x-5)(2x-1) (-2x+1)(-x+5)=0 and (-2x+1)(x-5)=0 quadratic equation have as solution set.
Answer:
well i got some HOPE THIS WILL HELP
Step-by-step explanation:
To find the solution set of a factored quadratic equation, you should set each one of the factors equal to zero and solve for .
Lets take our first quadratic equation (x+1/2)(x-5)=0 and apply this. First we are our factors equal to zero and ; next we are going to sole for in each factor to find the solution set:
and , so the solution set for the quadratic equation (x+1/2)(x-5)=0 is
Lets do the same for our next one (x-5)(2x-1)=0
and
So, the solution set for the quadratic equation (x-5)(2x-1)=0 is
Next one (x+5)(2x-1)=0
and
So, the solution set for the quadratic equation (x+5)(2x-1)=0 is
Next one (-2x+1)(-x+5)=0
and
So, the solution set for the quadratic equation (-2x+1)(-x+5)=0 is : therefore, we are going the select this one.
Next one (x+1/2)(x+5)=0
and
So, the solution set for the quadratic equation (x+1/2)(x+5)=0 is
Finally, our last one (-2x+1)(x-5)=0
and
So, the solution set for the quadratic equation (-2x+1)(x-5)=0 is ; 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 as solution set.
