This is my question 2x^2-8x=-25
I cant seem to figure out how to solve for x?
Add the -25 to the other side...
2x^2-8x+25=0,
Now use the quadratic formula to solve for x...
Hope this helps
I was told to start with
x^2-8x+16=7/2 so where do i go from there?
2x^2 - 8x =-25
divide each term by 2
using completing the square,
x^2 - 4x = -25/2
take 1/2 the coefficient of the x term, square it, then add it to each side
x^2 - 4x +4 = -25/2 +4
(x-2)^2 = -17/2 or -34/4 , (I anticipated to take √ of both sides)
x -2 = ± √-34/2 = ± i√34/2
x = 4/2 ± i√34/2 = (4 ± i√34)/2
or ... by the formula
2x^2 - 8x + 25 = 0
x = (8 ± √-136)/4
= (8 ± 2√-34)/2
= (4 ± i√34)/2
I was told to start with
x^2-8x+16=7/2 so where do i go from there? That's after / by 2.
To solve the equation 2x^2 - 8x = -25, we can follow these steps:
Step 1: Rewrite the equation in standard form
Rearrange the equation by moving all terms to one side, so we have:
2x^2 - 8x + 25 = 0
Step 2: Check if the equation can be factored
To determine if the equation is factorable, we'll calculate the discriminant (b^2 - 4ac) using the quadratic formula. If the discriminant is a perfect square, the equation can be factored. Otherwise, we'll proceed to the next step.
In our equation, a = 2, b = -8, and c = 25. Plugging these values into the quadratic formula (x = [-b ± √(b^2 - 4ac)] / 2a), we get:
Discriminant = (-8)^2 - 4(2)(25)
Discriminant = 64 - 200
Discriminant = -136
Since the discriminant is negative (-136), the equation cannot be factored.
Step 3: Solve using the quadratic formula
As we couldn't factor the equation, we'll use the quadratic formula to find the solutions for x:
x = [-b ± √(b^2 - 4ac)] / 2a
Plugging the values from our equation, we get:
x = [-(-8) ± √[(-8)^2 - 4(2)(25)]] / (2 * 2)
x = [8 ± √(64 - 200)] / 4
x = [8 ± √(-136)] / 4
Since the √(-136) is not a real number (due to the negative value under the square root), the original equation 2x^2 - 8x = -25 does not have any real solutions.
Therefore, there are no real values of x that satisfy the equation.