solve for x
(4/4)x^2-(12/4)x+(61/4)=0
I realize i need to do "complete the square" and here is what i got...
(4/4)x^2-(12/4)x+(61/4)=0
x^2-3x+15.25=0
x^2-3x=-15.25
x^2-3x+2.25= -15.25+2.25
(x-1.5)^2=-13
x=1.5 +/- sqrt(-13)
I know i am wrong... help?
Strange equation, I will assume you started with
x^2 - 3x + 15.25 = 0
or
x^2 - 3x + 61/4 = 0
or
4x^2 - 12x + 61 = 0
from: x^2 - 3x + 61/4 = 0
x^2 - 3x = -61/4
x^2 - 3x + 9/4 = -61/4 + 9/4
(x - 3/2)^2 = -13
x - 3/2 = ±√-13
x = 3/2 ± √-13
you had that, but are probably worried about the √-13
If you studied complex numbers you would finish by saying
x = 3/2 ±i√13
if you have not studied complex numbers, your conclusion is that the equation has no real solution.
Thank you so much!
To solve for x in the equation (4/4)x^2 - (12/4)x + (61/4) = 0, we can use the quadratic formula. The quadratic formula states that for any quadratic equation in the form ax^2 + bx + c = 0, the solutions for x are given by:
x = (-b ± √(b^2 - 4ac)) / 2a
In our equation, a = 4/4 = 1, b = -12/4 = -3, and c = 61/4.
Substituting these values into the quadratic formula, we have:
x = (-(-3) ± √((-3)^2 - 4(1)(61/4))) / (2(1))
Simplifying further, we get:
x = (3 ± √(9 - 4(61/4))) / 2
x = (3 ± √(9 - 61/1)) / 2
x = (3 ± √(-52/4)) / 2
Now, we can simplify the expression under the square root:
x = (3 ± √(-13)) / 2
Since the square root of a negative number is not a real number, this quadratic equation has no real solutions. Therefore, x has no solution in the real number system.