can someone show me the steps for this problem because i got it incorrect in my exam.
Solve each of the following quadratic equations.
4x^2=13x+12
what i had put was that x is 3/4 and x is -4
but i just got my result back and it was wrong...
4x^2 -13x _12 = 0
(4x+3) (x -4) = 0
x = 4 or -3/4
To solve the quadratic equation 4x^2 = 13x + 12, we need to follow these steps:
Step 1: Rewrite the equation in the standard form, which is ax^2 + bx + c = 0. In this case, we have 4x^2 - 13x - 12 = 0.
Step 2: Factorize the quadratic equation. Look for 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):
The factors of 4 are: 1, 2, 4.
The factors of -12 are: -1, -2, -3, -4, -6, -12.
After trying out different factors, we see that -4 and 3 multiply to give -12 and add up to -1. Therefore, we can rewrite the equation as:
(4x + 3)(x - 4) = 0.
Step 3: Set each factor equal to zero and solve for x:
Setting 4x + 3 = 0:
4x = -3
x = -3/4.
Setting x - 4 = 0:
x = 4.
Therefore, the solutions to the quadratic equation 4x^2 = 13x + 12 are x = -3/4 and x = 4.