I am still trying to solve for H in this equation but I am still lost
MgH = (1/2)*k*(H-2-10)^2 + Mg*2
I was told to expand the square on the right, gather terms, put it in the standard quadratic form, and use the quadratic formula, but I don't know how.
Should it be Mgh=(H-12)^2+25+1685?????
To solve for H in the equation MgH = (1/2)*k*(H-2-10)^2 + Mg*2, you need to follow a series of steps. Let's break it down:
Step 1: Expand the square on the right side of the equation
(H-2-10)^2 can be simplified as (H-12)^2 by combining the terms inside the parentheses. However, you have a mistake in your expansion. It should be: (H-12)^2 = H^2 - 24H + 144.
So, the equation becomes: MgH = (1/2)*k*(H^2 - 24H + 144) + Mg*2.
Step 2: Distribute the (1/2)*k to the terms inside the parentheses
Distribute (1/2)*k to each term: (1/2)*k*H^2 - 12kH + (1/2)*k*144.
The equation is now: MgH = (1/2)*k*H^2 - 12kH + 72k + Mg*2.
Step 3: Gather like terms
Group the terms with H: (1/2)*k*H^2 - MgH - 12kH = 72k + Mg*2.
The equation becomes: (1/2)*k*H^2 - (Mg + 12k)H = 72k + Mg*2.
Step 4: Put the equation in standard quadratic form
To achieve the standard quadratic form, move the constant terms to the right side: (1/2)*k*H^2 - (Mg + 12k)H - (72k + Mg*2) = 0.
Multiplying through by 2 to remove the fraction: k*H^2 - 2(Mg + 12k)H - 2(72k + Mg*2) = 0.
The equation is now: k*H^2 - 2(Mg + 12k)H - 2(72k + Mg*2) = 0.
Step 5: Use the quadratic formula to solve for H
The quadratic formula is: H = (-b ± √(b^2 - 4ac)) / 2a.
In this case, a = k, b = -2(Mg + 12k), and c = -2(72k + Mg*2).
Substitute the values into the quadratic formula and solve for H.
It's important to double-check your steps and calculations, as mistakes can easily occur during the process.