Question

Which of the following quadratic equations has the solution set (1/2,5)? Select all that apply

Answer 1

To find the solution set of a factored quadratic equation, you should set each one of the factors equal to zero and solve for $x$.

Lets take our first quadratic equation (x+1/2)(x-5)=0 and apply this. First we are our factors equal to zero $x+ \frac{1}{2} =0$ and $x-5=0$; next we are going to sole for $x$ in each factor to find the solution set:
$x=- \frac{1}{2}$ and $x=5$, so the solution set for the quadratic equation (x+1/2)(x-5)=0 is $(- \frac{1}{2},5)$

Lets do the same for our next one (x-5)(2x-1)=0
$x-5=0$
$x=5$
and
$2x-1=0$
$2x=1$
$x= \frac{1}{2}$
So, the solution set for the quadratic equation (x-5)(2x-1)=0 is $(5, \frac{1}{2})$

Next one (x+5)(2x-1)=0
$x+5=0$
$x=-5$
and 
$2x-1=0$
$2x=1$
$x= \frac{1}{2}$
So, the solution set for the quadratic equation (x+5)(2x-1)=0 is $(-5, \frac{1}{2} )$

Next one (-2x+1)(-x+5)=0
$-2x+1=0$
$-2x=-1$
$x= \frac{-1}{-2}$
$x= \frac{1}{2}$
and
$-x+5=0$
$-x=-5$
$x=5$
So, the solution set for the quadratic equation (-2x+1)(-x+5)=0 is $( \frac{1}{2} ,5)$: therefore, we are going the select this one.

Next one (x+1/2)(x+5)=0
$x+ \frac{1}{2} =0$
$x=- \frac{1}{2}$
and
$x+5=0$
$x=-5$
So, the solution set for the quadratic equation (x+1/2)(x+5)=0 is $(- \frac{1}{2} ,-5)$

Finally, our last one (-2x+1)(x-5)=0
$-2x+1=0$
$-2x=-1$
$x= \frac{-1}{-2}$
$x= \frac{1}{2}$
and
$x-5=0$
$x=5$
So, the solution set for the quadratic equation (-2x+1)(x-5)=0 is $( \frac{1}{2} ,5)$; 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 $( \frac{1}{2} ,5)$ as solution set.