Question

Let f(x)=`x^2`+2x-15. Determine the vaules of x for which f(x)=-12?

Answer 1

`x={-3, 1}`

Explanation:

Setting `f(x) = -12` gives us:

`-12=x^2+2x-15`

To solve quadratic equations, you need to set the equation equal to zero. By adding 12 to both sides, we get:

`0=x^2+2x-3`

From here, we can factor the quadratic to `0=(x+3)(x-1)`

Using the Zero Product Property, we can solve the equation by setting each factor equal to zero and solving for x.

`x+3=0 -> x=-3`

`x-1=0 -> x=1`

The two solutions are -3 and 1

Answer 2

x=-3 and x=1 .

Explanation:

Put f(x)=-12

`-12=x^2+2x-15`
`x^2+2x-15 +12=0`
`x^2+2x-3=0`

Time to factorize now
`x^2 + 3x -x -3=0`
`x(x+3) +(-1)(x+3)=0`

take x+3 common
`(x+3)(x-1)=0`

x=-3 and x=1 .

Answer 3

`1` or `-3`

Explanation:

Since `f(x)=-12`, then `x^2+2x-15=-12`. Solve this by factoring:

`x^2+2x-3=0`

`(x-1)*(x+3)=0`

`x-1=0`

`x+3=0`

The answer is

`x=1,-3`