I have to use my answer from a previous question (factorise and solve x^2+x-30=0 which i factorised as (x-5)(x+6)=0 and then solved as x=5 or x=-6) i now have to explain how i can use this answer to solve th equation 4x^2+4x-120=0 and then say what are the solutions?
Please could you help me!!!!!
Thank you
Do you see a numeric factor that is common in each term? Factor that number from each term. What is left?
Substitute each value in the second equation to see what fits.
Yes, I can help you with that!
To solve the equation 4x^2 + 4x - 120 = 0 using the previous answer, we can utilize the factored form of the equation x^2 + x - 30 = 0, which is (x - 5)(x + 6) = 0.
To begin, let's solve the equation (x - 5)(x + 6) = 0 by setting each factor to zero and solving for x:
Setting (x - 5) = 0:
x - 5 = 0
x = 5
Setting (x + 6) = 0:
x + 6 = 0
x = -6
Now, we can use these solutions for x in the equation 4x^2 + 4x - 120 = 0:
Substituting x = 5:
4(5)^2 + 4(5) - 120 = 0
100 + 20 - 120 = 0
0 = 0
Substituting x = -6:
4(-6)^2 + 4(-6) - 120 = 0
144 - 24 - 120 = 0
0 = 0
From the substitution, we can see that both values of x = 5 and x = -6 are valid solutions to the equation 4x^2 + 4x - 120 = 0. Thus, the solutions to this equation are x = 5 and x = -6.