SOLUTION: Please help me solve this equation
Solve the inequality.
(x - 5)(x^2 + x + 1) > 0
Algebra.Com
Question 1048804: Please help me solve this equation
Solve the inequality.
(x - 5)(x^2 + x + 1) > 0
Found 2 solutions by stanbon, Theo:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Please help me solve this equation
Solve the inequality.
(x - 5)(x^2 + x + 1) > 0
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The Real Number solutions are x > 5.
Cheers,
Stan H.
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Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
set the equation equal to 0.
you get (x-5) * (x^2 + x + 1) = 0
this is true if (x-5) = 0 and/or if (x^2 + x + 1) = 0
x-5 = 0 when x = 5.
this means that x-5 is negative when x < 5 and x-5 is positive when x > 5.
x^2 + x + 1 never equals 0.
in fact, it is always positive.
since x^2 + x + 1 = 0 is a quadratic equation in standard form, then it has a max/min point at x = -b/2a.
since a = 1 and b = 1 and c = 1, that max/min point will be at x = -1/2.
when x = -1/2, the max/min point will be at y = (-1/2)^2 - 1/2 + 1 which will be at y = 1/4 - 1/2 + 1 which becomes y = 3/4.
since the coefficient of the x^2 term is positive, then x = -1/2 is a min point and the equation opens up.
that says that all values of y = x^2 + x + 1 are positive regardless of the value of x.
not only are they all positive, but they are all greater than or equal to 3/4.
since x^2 + x + 1 is always positive and since x-5 is positive when x > 5 and negative when x < 5, then:
(x-5) * (x^2 + x + 1) is positive when x > 5 because positive times positive always gives a positiv
e result.
(x-5) * (x^2 + x + 1) is negative when x < 5 because positive times negative always give a negative result.
your solution is that (x-5) * (x^2 + x + 1) > 0 when x > 5.
the following graph confirms that this is true.
here's the graph of x-5 by itself.
here's the graph of x^2 + x + 1 by itself.
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