Question

How do you solve `-5x ^ { 2} - 16x = - 16`?

Answer 1

`x = -4 , 4/5`

Explanation:

Rearrange the equation like so, by sending RHS ( right hand side) to LHS ( left hand side) and multiplying both sides by -1:

`5x^2 + 16x - 16 = 0`

There are multiple ways to solve this.

But before we go about solving this equation, we have to make sure that a real solution exists. I'll explain how to check.

A general quadratic equation is of the form;

`ax^2 + bx + c = 0`

The easiest way to see if a real solution exists or not for such an equation is to use the relation;

`b^2 - 4ac >= 0`

If this relation is satisfied then a real solution exists.
And if its `> 0` then the equation has two different roots otherwise just one unique root exist.

On comparing we can see that;

`a = 5, b = 16 and c = -16`

Therefore,

`b^2 - 4ac = 16^2 - 4*5*(-16)`
`=> 256 + 320 = 576 >0`

Hence, the given equation has two different real roots.

Now, we can use the following formula to get the roots;

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

Using this formula we get the following solution;

`x = (-16 +- sqrt(576))/10`

`=> x = (-16 +- 24)/10`

`=>x = -4 , 4/5`

Answer 2

`x=-4" or "x = -4/5`

Explanation:

To solve a quadratic equation, first set it equal to `0`.

In this case moving the terms to the right will give us `+5x^2`

`0= 5x^2+16x-16`

Factorising: Find factors of `5 and 16` which subtract to give `16`

`color(white)(...........)5" "16`
`color(white)(..........)darr" "darr`
`color(white)(...........)5" "-4 rarr 1 xx -4 =" "-4`
`color(white)(...........)1" "+4 rarr 5 xx +4 =" " ul(+20)`
`color(white)(.....................................)"difference"=+16`

The factors are:

`(5x-4)(x+4)=0`

Setting each factor equal to 0 gives:

`x = 4/5 and x = -4` as solutions.