F(x)=x^2-2x-3
plz help me slove this ?
there's nothing to solve. You have just presented a function of x.
Now, if you want to find where f(x)=0, that's a problem you can solve. On the way, we'll discover how to factor the quadratic, which is another kind of problem.
F(x) = (x-3)(x+1)
so, F(x)=0 when
x-3 = 0
or
x+1 = 0
Those I think you can handle, eh?
OH OK THANK YOU
OH OK THANK YOU STEVE
To solve the equation f(x) = x^2 - 2x - 3, we need to find the values of x that satisfy this equation.
Step 1: Set the equation equal to zero.
x^2 - 2x - 3 = 0
Step 2: Factorize the quadratic expression if possible.
In this case, the expression cannot be easily factored. Therefore, we move to the next step.
Step 3: Use the quadratic formula.
The quadratic formula states:
For a quadratic equation in the form ax^2 + bx + c = 0, the solutions (or roots) can be found using the formula:
x = (-b ± √(b^2 - 4ac))/(2a)
For the equation x^2 - 2x - 3 = 0, a = 1, b = -2, c = -3.
x = (-(-2) ± √((-2)^2 - 4 *1 * -3))/(2*1)
x = (2 ± √(4 + 12))/2
x = (2 ± √(16))/2
x = (2 ± 4)/2
This gives us two potential solutions: x = (2 + 4)/2 and x = (2 - 4)/2.
Simplifying further:
x = 6/2 and x = -2/2
This leads to the solutions:
x = 3 and x = -1.
So, the solutions to the equation f(x) = x^2 - 2x - 3 are x = 3 and x = -1.