Asked by Kim
How do you determine whether/how many nonreal solutions an equation has?
The statement:
x^3 - 4x^2 - 3x + 2 = 0 has two nonreal solutions is false
How do I solve this?
The statement:
x^3 - 4x^2 - 3x + 2 = 0 has two nonreal solutions is false
How do I solve this?
Answers
Answered by
Reiny
Have you studied the Descarte's Rule of Signs ?
here is a good page for it
http://www.purplemath.com/modules/drofsign.htm
or else Google the title for other sources.
or
A quick sketch will show that it crosses the x-axis 3 times.
So there are 3 real solutions.
or
A quick test for x = ±1 and ±2 shows that x = -1 is a solution
so x+1 will be a factor,
By synthetic division
x^3 - 4x^2 - 3x + 2 = (x+1)(x^2 - 5x + 2)
solving x^2 - 5x + 2 = 0
x = (5 ± √17)/2
so x = -1, (5 ± √17)/2 which shows it as 3 real roots.
here is a good page for it
http://www.purplemath.com/modules/drofsign.htm
or else Google the title for other sources.
or
A quick sketch will show that it crosses the x-axis 3 times.
So there are 3 real solutions.
or
A quick test for x = ±1 and ±2 shows that x = -1 is a solution
so x+1 will be a factor,
By synthetic division
x^3 - 4x^2 - 3x + 2 = (x+1)(x^2 - 5x + 2)
solving x^2 - 5x + 2 = 0
x = (5 ± √17)/2
so x = -1, (5 ± √17)/2 which shows it as 3 real roots.
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