Question

How do you solve `4x^2 - 5x=0` using the quadratic formula?

Answer 1

`x=0 or x=5/4`

Explanation:

The quadratic formula for `ax^2+bx+c=0` is given by `x=(-b+-sqrt(b^2-4ac))/(2a)`

`a=4, b=-5, c=0`

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

`x=(5+-sqrt(25))/8`

`x=(5+-5)/8=>x=0 or x=10/8=5/4`

Answer 2

`x=5/4 or x=0`

Explanation:

The equation `y=4x^2-5x=0` is written in the form `y=ax^2+bx+c`,
so
`a=4`, `b=-5`, `c=0`

The quadratic formula is `x=(-b+-sqrt(b^2-4ac))/(2a)`

Substitute the values of a, b and c into the formula

`x=(5+-sqrt(25))/(8)`

`x=(5+sqrt(25))/(8)` or `x=(5-sqrt(25))/(8)`

`x=(10)/(8)` or `x=0/(8)`

`x=5/4 or x=0`

Answer 3

`x=0,5/4`

Explanation:

`4x^2-5x=0` is a quadratic equation in standard form:

`ax^2+bx+c=0`,

where:

`a=4`, `b=-5`, `c=0`

Quadratic Formula

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

Plug in the known values and solve.

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

Simplify.

`x=(5+-sqrt25)/8`

`x=(5+-5)/8`

`x=(5+5)/8=10/8=5/4`

`x=(5-5)/8=0/8=0`

`x=0,5/4`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`4x^2-5x=0` can also be solved by factoring.

Factor out the common `x`.

`x(4x-5)=0`

`x=0`

`4x-5=0`

`4x=5`

`x=5/4`

`x=0,5/4`