Please help!
I need to factorise and solve this equation,
x^2 +x -30 =0
I then need to explain how I would use the answer drom this to solve the equation
4x^2 +4x-120=0
first one ...
(x+6)(x-5) = 0
x = -6 or x = 5
2nd
4x^2 +4x-120=0 divide each term by 4
x^2 + x - 30 = 0
mmmhh!!
Many thanks for your help, I may be being a bit slow but how would I use the answer from the first equation to solve the second one?
To factorize and solve the equation x^2 + x - 30 = 0, we can use a quadratic factoring method. Here are the steps:
Step 1: Look at the quadratic equation and identify the values of a, b, and c in the form ax^2 + bx + c = 0.
In this case, a = 1, b = 1, and c = -30.
Step 2: Find the product of a and c, that is, ac = 1 * -30 = -30.
Step 3: Think of two numbers whose product is equal to ac = -30 and whose sum is equal to b = 1. In this case, the numbers are 6 and -5 because 6 * -5 = -30 and 6 + (-5) = 1.
Step 4: Rewrite the expression with these numbers:
x^2 + 6x - 5x - 30 = 0.
Step 5: Factor by grouping:
(x^2 + 6x) - (5x + 30) = 0.
x(x + 6) - 5(x + 6) = 0.
Step 6: Factor out the common term (x + 6):
(x - 5)(x + 6) = 0.
Step 7: Set each factor equal to zero and solve for x:
x - 5 = 0 → x = 5.
x + 6 = 0 → x = -6.
Therefore, the solutions to the equation x^2 + x - 30 = 0 are x = 5 and x = -6.
Now, let's move on to the second equation, 4x^2 + 4x - 120 = 0.
Step 1: Divide the equation by the common factor, which is 4:
x^2 + x - 30 = 0, which we already solved.
Step 2: Rewrite the equation using the solutions from the previous equation:
(x - 5)(x + 6) = 0.
Step 3: Solve for x by setting each factor equal to zero:
x - 5 = 0 → x = 5.
x + 6 = 0 → x = -6.
So, the solutions to the equation 4x^2 + 4x - 120 = 0 are x = 5 and x = -6.