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Real root question watch

1. The polynomial f(x) is given by f(x) = x^3 + 4x - 5 OR (x - 1) (x^2 + x + 5)

Show that the equation f(x) = 0 has exactly one real root and state its vale

two questions
1) not just here, but for any question like this can I only use equations where x has a power of 2 (not cubic powers!)

2)
If I use cubic powers b^2 - 4ac = 16 + 20 = 36
If I use the square from (x - 1) (x^2 + x + 5) then I get
b^2 - 4ac = 1 - 20 = -19

so I get no real roots whatsoever for b^2 - 4ac = 0

Whats going on
2. (Original post by RCB)
The polynomial f(x) is given by f(x) = x^3 + 4x - 5 OR (x - 1) (x^2 + x + 5)

Show that the equation f(x) = 0 has exactly one real root and state its vale

two questions
1) not just here, but for any question like this can I only use equations where x has a power of 2 (not cubic powers!)

2)
If I use cubic powers b^2 - 4ac = 16 + 20 = 36
If I use the square from (x - 1) (x^2 + x + 5) then I get
b^2 - 4ac = 1 - 20 = -19

so I get no real roots whatsoever for b^2 - 4ac = 0

Whats going on
The key is to use (x - 1) (x^2 + x + 5), for the (x - 1), we can clearly see that it has a solution of x=1, as you have found out (x^2 + x + 5) has no real roots, so we can see the cubic has only 1 real root. You can't use the discriminant for the whole cubic, as that's only for quadratics.
3. Well you know that one root is going to be at x=1, if you can show that the quadratic will not have any roots you have done it.
4. (x - 1) (x^2 + x + 5) = 0 tells you that x=1 is a real root.

b^2 - 4ac applies to quadratics only, so x^2 + x + 5 has a discriminant of -19<0 so yields no further real solutions.

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Updated: March 26, 2011
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