Question

How do you solve `2x^2+5x=0` using the quadratic formula?

Answer 1

`x={0,-5/2}`

Explanation:

`ax^2+bx+c=0`

`2x^2+5x=0`

`a=2" ; "b=5" ; "c=0`

`Delta=sqrt(b^2-4*a*c)`

`Delta=sqrt(5^2-4*2*0)`

`Delta=sqrt 5^2`

`Delta=±5`

`x_("1,2")=(-b±Delta)/(2*a)`

`x_1=(-5-5)/(2*2)=-10/4=-5/2`

`x_2=(-5+5)/(2*2)=0`

`x={0,-5/2}`

`"Or simply"`

`2x^2+5x=0`

`2cancel(x^2)=-5cancel(x)`

`2x=-5`

`x=-5/2`

Answer 2

`x=-5/2, 0`

Explanation:

`2x^2+5x=0` is a quadratic equation in the form `ax^2+bx+c`, where `a=2`, `5=b`, and `c=0`.

Quadratic formula

`x=(-b+-sqrt(b^2-4ac))/(2a)`

Plug the given values into the formula.

`x=(-5+-sqrt((-5)^2-4*2*0))/(2*2)`

Simplify.

`x=(-5+-sqrt(25-0))/4`

Simplify.

`x=(-5+-sqrt 25)/4`

Simplify.

`x=(-5+-5)/4`

Solve for `x_1` and `x_2`.

`color(blue)(x_1=(-5+5)/4)`

`color(blue)(x_1=0/4)`

`color(blue)(x_1=0)`

`color(red)(x_2=(-5-5)/4)`

`color(red)(x_2=-10/4)`

`color(red)(x_2=-5/2)`

`color(purple)(x=-5/2, 0)`