How do you find a quadratic equation if you are given solutions such as x=2/3 and x=-4
(x-2/3)(x+4) = 0
(3x-2)(x+4) = 0
3x^2 + 10 x - 8 = 0
To find a quadratic equation with given solutions, we can use the fact that the solutions (also known as the roots) of a quadratic equation can be expressed using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a.
In this case, you are given the solutions x = 2/3 and x = -4. We can set up two separate equations using these known solutions:
Equation 1:
When x = 2/3,
0 = a(2/3)^2 + b(2/3) + c
Equation 2:
When x = -4,
0 = a(-4)^2 + b(-4) + c
Now we can solve this system of equations simultaneously to find the values of a, b, and c.
1. Equation 1:
0 = (4a/9) + (2b/3) + c
Multiply through by 9 to clear the fraction:
0 = 4a + 6b + 9c
2. Equation 2:
0 = 16a - 4b + c
Now we have a system of linear equations:
4a + 6b + 9c = 0 [Equation 1]
16a - 4b + c = 0 [Equation 2]
Solve this system of equations to determine the values of a, b, and c.