How Many Real Roots Does This Equation Have?

  • MHB
  • Thread starter anemone
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In summary, the equation for finding the number of real roots in a quadratic equation is given by the discriminant, which is b^2 - 4ac. The significance of the number of real roots in an equation is that it determines the number of solutions, shape of the graph, and behavior of the equation. A quadratic equation can have a maximum of two real roots, but it can also have complex roots. The number of real roots can also help in choosing the appropriate method for solving the equation.
  • #1
anemone
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MHB
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Here is this week's POTW:

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Find the number of real roots of the equation $1+\dfrac{x}{1!}+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\dfrac{x^5}{5!}+\dfrac{x^6}{6!}=0$.

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  • #2
Congratulations to the following members for their correct solution!(Cool)

1. lfdahl
2. Opalg

Solution from Opalg:
Let $P(x) = 1+\dfrac{x}{1!}+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\dfrac{x^5}{5!}+\dfrac{x^6}{6!}$. Then
$$\begin{aligned}720P(x) &= x^6 + 6x^5 + 30x^4 + 120x^3 + 360x^2 + 720x + 720 \\ &= (x^3 + 3x^2 + 6x + 15)^2 + 9x^4 + 54x^3 + 234x^2 + 540x + 495 \\ &= (x^3 + 3x^2 + 6x + 15)^2 + 9(x^4 + 6x^3 + 26x^2 + 60x + 55) \\ &= (x^3 + 3x^2 + 6x + 15)^2 + 9\bigl((x^2 + 3x + 4)^2 + 9x^2 + 36x + 39\bigr) \\ &= (x^3 + 3x^2 + 6x + 15)^2 + 9\bigl((x^2 + 3x + 4)^2 + 9(x + 2)^2 + 3\bigr) \\ &= (x^3 + 3x^2 + 6x + 15)^2 + 9(x^2 + 3x + 4)^2 + 81(x + 2)^2 + 27.\end{aligned}$$ That is a sum of squares, so it is always positive (with minimum value at least $27$). Therefore $P(x)$ is always positive and so the equation $P(x) = 0$ has no real roots.
 

Related to How Many Real Roots Does This Equation Have?

1. How do you determine the number of real roots of an equation?

To determine the number of real roots of an equation, you can use the discriminant formula. This formula is b^2 - 4ac, where a, b, and c are the coefficients of the equation. If the discriminant is greater than zero, the equation has two real roots. If the discriminant is equal to zero, the equation has one real root. If the discriminant is less than zero, the equation has no real roots.

2. Can an equation have more than two real roots?

No, an equation can only have a maximum of two real roots. This is because the fundamental theorem of algebra states that a polynomial equation of degree n can have at most n complex roots, including both real and imaginary roots. Since a real root is a subset of complex roots, an equation can have at most n real roots, which is two for a quadratic equation.

3. What is the difference between real and imaginary roots?

Real roots are solutions to an equation that are real numbers. They can be graphed on a number line and have a physical meaning. Imaginary roots are solutions to an equation that involve the imaginary unit, i = √-1. They cannot be graphed on a number line and do not have a physical meaning.

4. Can an equation have both real and imaginary roots?

Yes, an equation can have both real and imaginary roots. This is because the solutions to a polynomial equation can be both real and imaginary numbers. For example, a quadratic equation with a negative discriminant will have two imaginary roots.

5. How do you use the quadratic formula to find the number of real roots of an equation?

The quadratic formula, x = (-b ± √(b^2 - 4ac)) / 2a, can be used to find the number of real roots of an equation. If the discriminant, b^2 - 4ac, is positive, the equation will have two real roots. If the discriminant is zero, the equation will have one real root. If the discriminant is negative, the equation will have no real roots. This is because the square root of a negative number will result in an imaginary number.

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