Asked by confused
solve the equation in the real number system.
2x^4-23x^3+75x^2-88x+28=0
What are the real solutions of the equations?
please show work. I do not know how to work this at all
2x^4-23x^3+75x^2-88x+28=0
What are the real solutions of the equations?
please show work. I do not know how to work this at all
Answers
Answered by
Damon
x f(x)
1 -6
0 28
-1 216
so there is a zero between zero x=0 AND x = 1
You can iterate and find it
then divide by (x-xo)
However before doing that I would search on Google for a polynomial grapher.
1 -6
0 28
-1 216
so there is a zero between zero x=0 AND x = 1
You can iterate and find it
then divide by (x-xo)
However before doing that I would search on Google for a polynomial grapher.
Answered by
Damon
http://mathportal.org/calculators/polynomials-solvers/polynomial-graphing-calculator.php
yields 1/2 , 2 , 7
yields 1/2 , 2 , 7
Answered by
Damon
by the way you can see from the graph that 2 is a double root
If you were clever enough to factor this you would have
(2x-1)(x-2)(x-2)(x-7)
If you were clever enough to factor this you would have
(2x-1)(x-2)(x-2)(x-7)
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