Problem 17 - Algebra and Trigonometry

Question

\(5-60\) Find all real solutions of the equation. $$ x^{3}-x^{2}+x-1=x^{2}+1 $$

Step-by-Step Solution

Answer

The only real solution is \(x = 2\).

Step 1: Set the Equation to Zero

Start by moving all the terms to one side of the equation to set it to zero. Subtract each term on the right-hand side from the left-hand side: \[x^3 - x^2 + x - 1 - x^2 - 1 = 0\] Simplify this to: \[x^3 - 2x^2 + x - 2 = 0\]

Step 2: Identify Possible Rational Roots

Use the Rational Root Theorem to identify possible rational roots of the polynomial \(x^3 - 2x^2 + x - 2\). The potential rational roots are the factors of the constant term over the factors of the leading coefficient. In this case, they are \(\pm 1, \pm 2\).

Step 3: Test Possible Roots

Test the possible rational roots by substituting them into the polynomial: - For \(x = 1\): \[1^3 - 2\times1^2 + 1 - 2 = 1 - 2 + 1 - 2 = -2\] (Not a root) - For \(x = -1\): \[(-1)^3 - 2(-1)^2 + (-1) - 2 = -1 - 2 - 1 - 2 = -6\] (Not a root) - For \(x = 2\): \[2^3 - 2\times2^2 + 2 - 2 = 8 - 8 + 2 - 2 = 0\] (2 is a root)

Step 4: Use Synthetic Division

Perform synthetic division using the root \(x = 2\) to factor the polynomial. The coefficients of the polynomial \(x^3 - 2x^2 + x - 2\) are \(1, -2, 1, -2\). Write them as row \([1, -2, 1, -2]\) and divide using \(x = 2\): \[\begin{array}{r|rrrr}2 & 1 & -2 & 1 & -2 \ & & 2 & 0 & 2 \\hline & 1 & 0 & 1 & 0 \\end{array}\] The quotient is \(x^2 + 1\), and the remainder is 0. This confirms \(x = 2\) is a root. The polynomial factors to \((x - 2)(x^2 + 1)\).

Step 5: Solve the Quadratic Equation

Now solve \(x^2 + 1 = 0\) for real roots: \[x^2 + 1 = 0\] \[x^2 = -1\] This equation has no real solutions since the square of a real number cannot be negative.

Step 6: Identify All Real Solutions

The only real solution to the equation \(x^3 - 2x^2 + x - 2 = 0\) is \(x = 2\).

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a useful tool for finding potential rational roots of polynomial equations. It states that any rational solution, in the form of \(\frac{p}{q}\), must have \(p\) as a factor of the constant term and \(q\) as a factor of the leading coefficient. For the polynomial \(x^3 - 2x^2 + x - 2\), this theorem helps us identify possible rational roots as \(\pm 1\) and \(\pm 2\).
First, list the factors of the constant term (which is -2): these are \(-1, +1, -2, +2\). Next, look at the factors of the leading coefficient (which is 1): these are \(+1\) and \(-1\). Therefore, by combining these, the only possibilities are \(\pm 1\) and \(\pm 2\).
Testing these possible roots by substituting them into the polynomial determines which ones are actual roots. If substituting a root makes the polynomial equal zero, it confirms that it is indeed a rational root.

Synthetic Division

Synthetic division is a simplified and efficient method for dividing a polynomial by a linear binomial (such as \(x - c\)). It's especially helpful when testing potential roots identified by the Rational Root Theorem. Let's see how it works using the polynomial \(x^3 - 2x^2 + x - 2\) and the root \(x = 2\).
Write down the coefficients of the polynomial \([1, -2, 1, -2]\). Then, perform synthetic division with \(x = 2\):
- Start by bringing down the leading coefficient, which is 1.
- Multiply it by the root 2, and write the result (2) under the next coefficient (-2).
- Add these numbers to get 0. Repeat this process until reaching the end of the coefficients list.
The result \([1, 0, 1, 0]\) shows a remainder of 0, confirming that \(x = 2\) is indeed a root. The process divides the polynomial into \((x - 2)(x^2 + 1)\), allowing us to focus on simpler factors.

Quadratic Equation

Quadratic equations usually take the form \(ax^2 + bx + c = 0\), which allows for various solving methods depending on the situation. When the factor of a cubic equation becomes a quadratic equation, as in \(x^2 + 1\), finding real roots might coincide with understanding its properties.
In this case, we're exploring whether \(x^2 + 1 = 0\) has real solutions. Rearranging gives us \(x^2 = -1\). However, no real number squared gives a negative result, confirming that there are no real solutions for this quadratic: it has complex roots, \(x = i\) and \(x = -i\).
In the context of our original equation \(x^3 - 2x^2 + x - 2 = 0\), this means the only real solution is the root we previously identified: \(x = 2\). Quadratic equations often simplify problems by narrowing possibilities and confirming solution types.