McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
9. Perfect Squares
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Exercise 56 Page 70

Look for a quadratic equation that has no real solutions and multiply both sides of it by x.

Example Equation: x^3+x = 0

Practice makes perfect
We have to find a counterexample to the following statement. A polynomial equation of degree three always has three real solutions. We need to find a polynomial of degree three that has less than three solutions. We can start by considering a quadratic equation that has no real solutions. Below we write one example equation.

x^2+1 = 0 There is no real solution to the equation above. However, it is a quadratic equation. We can obtain a third degree equation by multiplying it by x. x(x^2+1) = x(0) ⇒ x^3+x = 0 Notice that the final equation can be factored as x(x^2+1)=0. Then, by the Zero Product Property we can set the following two equations. x(x^2+1) = 0 ↙ ↘ x=0 x^2+1 = 0 Since the right-hand equation has no solution, we can conclude that the unique solution to the third degree equation is x=0. Therefore, this is the counterexample we were asked for.

Polynomial Equation Nº of Solutions
x^3+x = 0 1
Keep in mind that this is just an example equation and so your answer may vary.

Extra

Extra
To see why x^2+1=0 has no real solutions, let's solve this equation for x.
x^2+1 = 0
x^2 = -1
x = ± sqrt(-1)
Since sqrt(-1) is not a real number, we conclude that the equation x^2+1=0 has no real solutions.