Solve each of the following quadratic equations.
can some explain how to do these kinds of problems
4x^2=13x+12
i believe if i am understanding this i am supose to add to where i get =0
4x^2-13x+12=0
now what do i need to do
I am thinking factoring but I am not sure how to do this part
your equation should have been
4x^2 - 13x - 12 = 0
It does factor, I will give you a hint:
one of the factors is x-4
You must have been taught at least one general method to factor those quadratics that do factor.
It has been a while since i worked with any algebra
but i think i can remember some so this is what i got see if i am correct please
4x^2-13x-12=0
(4x+3)(x-4)=0
4x+3=0 0r x-4=0
4x=-3
4x/4=-3/4
x=-3/4
x-4=0
x=4
(-3/4,4)
did i do this right
yes, good for you.
it seems to be correct
Yes, you have solved the quadratic equation correctly. The factoring method you used is one of the ways to solve quadratic equations. Here's a step-by-step explanation:
1. Start with the original equation: 4x^2 - 13x - 12 = 0
2. Now, you need to find two numbers whose product is equal to the product of the coefficient of x^2 (which is 4) and the constant term (which is -12), and whose sum is equal to the coefficient of x (which is -13).
3. In this case, the numbers are -12 and 1, because (-12) * (1) = -12 and (-12) + (1) = -13.
4. Write the equation using these numbers as the coefficients of x. It becomes: 4x^2 - 12x + x - 12 = 0
5. Group the terms so that you can factor by grouping: (4x^2 - 12x) + (x - 12) = 0
6. Factor out common terms from each group: 4x(x - 3) + 1(x - 3) = 0
7. You will notice that (x - 3) is a common factor in both terms. Factor it out: (x - 3)(4x + 1) = 0
8. Set each factor equal to zero and solve for x: x - 3 = 0 or 4x + 1 = 0
9. Solve the first equation: x - 3 = 0. Add 3 to both sides: x = 3
10. Solve the second equation: 4x + 1 = 0. Subtract 1 from both sides: 4x = -1. Divide by 4: x = -1/4
So the solutions to the quadratic equation 4x^2 - 13x - 12 = 0 are x = 3 and x = -1/4.