SOLUTION: I am ridiculously confused on these problems. I don't even know where to start with them.
x²+41 =8x and the answer has to be in the form of a+bi
The second one is:
x²+13/2x
Algebra.Com
Question 571908: I am ridiculously confused on these problems. I don't even know where to start with them.
x²+41 =8x and the answer has to be in the form of a+bi
The second one is:
x²+13/2x=3
and then the last one is:
The width of a rectangle is 1 ft less than the length. The area is 2ft²
width=
length=
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
x²+41 =8x and the answer has to be in the form of a+bi
---
x^2 -8x + 41 = 0
x = [8 +- sqrt(64-4*41)]/2
x = [8 +- sqrt(-100)]/2
---
x = 4 +- 5i
=========================
The second one is:
x²+13/2x=3
x^2 + 13 -6x = 0
x^2 -6x + 13 = 0
x = [6 +- sqrt(36-4*13)]/2
x = [6 +- sqrt(-16)]/2
x = 3 +- 2i
=========================
and then the last one is:
The width of a rectangle is 1 ft less than the length. The area is 2ft²
Let length be "x" ; then width x-1
Equation:
x(x-1) = 2
x^2-x-2 = 0
Positive solution:
(x-2)(x+1) = 0
x = 2 (length)
x-1 = 1 (width)
======================
Cheers,
Stan H.
======================
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