Solve for x.
x(x+3)=2^2
I think I'm doing this all wrong. Here is the work I've done so far:
x^2+3x=4
x^2+3x-4=0
...now what?
i started trying to use the equation x=-b sqrt of (b^2-4ac)
but I'm just getting confused...
repeat
x^2+3x-4=0 OK, now:
(x+4)(x-1) = 0
x = 1 or x = -4
or
x = [-3 +/- sqrt (9+16) ] / 2
= -3/2 +/- 5/2
= -1.5 +/- 2.5
= -4 or +1 again
To solve this equation, we will use a method called factoring. Here's what you need to do:
Step 1: Rewrite the equation in standard form.
x^2 + 3x - 4 = 0
Step 2: Identify two numbers whose product is -4 (the coefficient of x^2) and whose sum is 3 (the coefficient of x). In this case, the numbers are -1 and 4.
Step 3: Split the middle term using these numbers.
x^2 + 4x - x - 4 = 0
Step 4: Factor by grouping.
(x^2 + 4x) - (x + 4) = 0
Step 5: Factor out the common terms.
x(x + 4) - 1(x + 4) = 0
Step 6: We now have a common binomial factor.
(x - 1)(x + 4) = 0
Step 7: Set each factor equal to zero and solve for x.
x - 1 = 0 or x + 4 = 0
x = 1 or x = -4
Therefore, the solutions to the equation x(x + 3) = 2^2 are x = 1 and x = -4.