Help Algebra 2/Trigonometry Problem

[daodude1987]

I'm having trouble with proving this equation: x=-b\2a
I am not really familiar with this equation but my trigonometry teacher says it is something from Algebra 2. How can I prove why or how this equation works for finding the x-coordinate?

 

[iggybaseball]

Taking into account that its almost midnight, i think i can scrap together an answer.

you are talking about using a specific equation:

ax^2 + bx + c ( quadratic)

however you use this equation:

f(x) = a(x-h)^2 + k (to put it into an equation that will make it easier to find the vertex of the equation)

h = -b/2a
k = c - ah^2 ( I am almost positive this is right)

the vertex will equal (h,k) so if the vertex liex on the x-axis u will have an x coordinate of the graph. hope that helps some

 

[M98Ranger]

To prove this equation, we need to use the quadratic formula: x = (-b ± √(b²-4ac)) / 2a. This formula is used to find the roots of a quadratic equation in the form of ax² + bx + c = 0.

Now, let's compare this formula with the given equation x = -b/2a. We can see that the only difference is the absence of the square root term in the given equation.

To understand why this equation works for finding the x-coordinate, we need to look at the discriminant (b²-4ac) in the quadratic formula. The discriminant determines the nature of the roots of a quadratic equation. If the discriminant is positive, the equation has two distinct real roots, if it is zero, the equation has one real root, and if it is negative, the equation has two complex roots.

In the given equation x = -b/2a, we can see that the discriminant is not present. This means that the equation has only one real root, which is the x-coordinate. This is because the quadratic formula gives us the two possible values for x, but in this case, the ±√(b²-4ac) term becomes zero, leaving us with only one value for x, which is the x-coordinate.

To further prove this, we can substitute the value of x = -b/2a in the original quadratic equation ax² + bx + c = 0. We will get (-b/2a)(a(-b/2a)² + b(-b/2a) + c) = 0, which simplifies to (-b/2a)(b²/4a²) = 0. This gives us the same result as x = (-b ± √(b²-4ac)) / 2a, but in this case, the ±√(b²-4ac) term becomes zero, leaving us with only one value for x, which is the x-coordinate.

Therefore, we can conclude that the given equation x = -b/2a works for finding the x-coordinate because it is a simplified version of the quadratic formula, specifically for equations with only one real root.