I need help to set the equation up to solve a system to ensure that I get the answer of (-2, 4)? I also need tosolve two problems in the quadratic equation method, the two problems are x to the second power - 2x - 13 = 0 then the other is x to the second power + 12x - 64 = 0 I need to know the set up of these three problems. please help me.
To set up a system of equations to solve for a specific solution, in this case (-2, 4), you need to create a set of equations where the given solution satisfies both equations. Here's an example of setting up a system:
Let's consider a system of linear equations:
1) Equation 1: 2x + y = -6
2) Equation 2: 3x - y = 16
To check if (-2, 4) is a solution, substitute x = -2 and y = 4 into both equations:
1) Equation 1: 2(-2) + 4 = -6 -> -4 + 4 = -6 -> 0 = -6 (not true)
2) Equation 2: 3(-2) - 4 = 16 -> -6 - 4 = 16 -> -10 = 16 (not true)
Since (-2, 4) does not satisfy both equations, it is not the solution to this system.
Now, let's solve the two quadratic equations:
1) x^2 - 2x - 13 = 0:
To solve this equation, you can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a)
Here, a = 1, b = -2, and c = -13.
Substituting the values, the quadratic formula becomes: x = (-(-2) ± √((-2)^2 - 4(1)(-13))) / (2(1))
Simplifying further: x = (2 ± √(4 + 52)) / 2
x = (2 ± √(56)) / 2
x = (2 ± 2√14) / 2
Simplifying: x = 1 ± √14
Therefore, the solutions to the equation x^2 - 2x - 13 = 0 are: x = 1 + √14 and x = 1 - √14.
2) x^2 + 12x - 64 = 0:
Again, we use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a)
Here, a = 1, b = 12, and c = -64.
Applying the quadratic formula: x = (-(12) ± √((12)^2 - 4(1)(-64))) / (2(1))
Simplifying: x = (-12 ± √(144 + 256)) / 2
x = (-12 ± √(400)) / 2
x = (-12 ± 20) / 2
Simplifying further: x = -6 ± 10
Therefore, the solutions to the equation x^2 + 12x - 64 = 0 are: x = -6 + 10 and x = -6 - 10.