Question

For the function f(x) = x^2 + 4x -5 solve the following f(x)=0

Answer 1

That's a question about quadratic function.

Any quadratic function can be represented by the following form:

`\boxed{f(x)=ax^2+bx+c}`

Example:

`f(x)= -3x^2-9x+57` is a function where
`a=-3`,
`b=-9` and
`c=57`.

Okay, in our problem, we need to find the value of x when
`f(x)=0`. That's mean that the result of our function is equal to zero. Therefore, we have the quadratic equation below:

`x^2+4x-5=0`

To solve a quadratic equation, we use the Bhaskara's formula. Do you remember the value of a, b and c? They going to be important right now. This is the Bhaskara's formula:

`\boxed{x=(-b\pm √(b^2-4ac) )/(2a) }`

So, let's see the values of a, b and c in our equation and apply them in the Bhaskara's formula:

In
`x^2+4x-5=0` equation,
`a=1`,
`b=4` and
`c=-5`. Let's replace those values:

`x=(-b\pm √(b^2-4ac) )/(2a)`

`x=(-4\pm √(4^2-4*1*(-5)) )/(2\cdot1)`

`x=(-4\pm √(16-(-20)) )/(2)`

`x=(-4\pm √(16 + 20)) )/(2)`

`x=(-4\pm √(36) )/(2)`

`x=(-4\pm 6 )/(2)`

From now, we have two possibilities:

To add:

`x_1 = (-4+6)/(2) \\x_1=(2)/(2) \\x_1=1`

To subtract:

`x_2=(-4-6)/(2) \\x_2=(-10)/(2) \\x_2=-5`

Therefore, the result of our problem is:
`x_1 = 1` and
`x_2=-5`.

I hope I've helped. ^^

Enjoy your studies. \o/