Question

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

(x+1/2)(x-5)=0
(x-5)(2x-1)=0
(x+5)(2x-1)=0
(-2x+1)(-x+5)=0
(x+1/2)(x+5)=0
(-2x+1)(x-5)=0

Answer 1

B, D, F

Explanation:

credit to person above,

Hope it helps!

Answer 2

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+ (1)/(2) =0` and
`x-5=0`; next we are going to sole for
`x` in each factor to find the solution set:

`x=- (1)/(2)` and
`x=5`, so the solution set for the quadratic equation (x+1/2)(x-5)=0 is
`(- (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= (1)/(2)`
So, the solution set for the quadratic equation (x-5)(2x-1)=0 is
`(5, (1)/(2))`

Next one (x+5)(2x-1)=0

`x+5=0`

`x=-5`
and

`2x-1=0`

`2x=1`

`x= (1)/(2)`
So, the solution set for the quadratic equation (x+5)(2x-1)=0 is
`(-5, (1)/(2) )`

Next one (-2x+1)(-x+5)=0

`-2x+1=0`

`-2x=-1`

`x= (-1)/(-2)`

`x= (1)/(2)`
and

`-x+5=0`

`-x=-5`

`x=5`
So, the solution set for the quadratic equation (-2x+1)(-x+5)=0 is
`( (1)/(2) ,5)`: therefore, we are going the select this one.

Next one (x+1/2)(x+5)=0

`x+ (1)/(2) =0`

`x=- (1)/(2)`
and

`x+5=0`

`x=-5`
So, the solution set for the quadratic equation (x+1/2)(x+5)=0 is
`(- (1)/(2) ,-5)`

Finally, our last one (-2x+1)(x-5)=0

`-2x+1=0`

`-2x=-1`

`x= (-1)/(-2)`

`x= (1)/(2)`
and

`x-5=0`

`x=5`
So, the solution set for the quadratic equation (-2x+1)(x-5)=0 is
`( (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
`( (1)/(2) ,5)` as solution set.