Problem 31 - Basic Technical Mathematics with Calculus

Question

Solve the given problems. Equations of the form \(y^{2}=x^{3}+a x+b\) are called elliptic curves and are used in cryptography. If \(y=3\) for the curve \(y^{2}=x^{3}-4 x+6,\) use synthetic division to show that one possible value of \(x\) is \(x=-1\). Then find any other possible values of \(x\).

Step-by-Step Solution

Answer

Possible values of \(x\) are \(-1\), \(\frac{1 + \sqrt{13}}{2}\), and \(\frac{1 - \sqrt{13}}{2}\).

Step 1: Use Given Values in the Elliptic Equation

We start with the elliptic curve equation given by \(y^2 = x^3 - 4x + 6\). We are also given that \(y=3\). Substitute the given \(y\) value into the equation:\[3^2 = x^3 - 4x + 6\] This simplifies to: \[9 = x^3 - 4x + 6\] Rearrange this to form a polynomial equation:\[x^3 - 4x - 3 = 0\]

Step 2: Verify x = -1 Using Synthetic Division

To prove that \(x = -1\) is a solution, apply synthetic division. We divide \(x^3 - 4x - 3\) by \((x + 1)\).Setup:- Coefficients of the polynomial are \(1, 0, -4, -3\) since there is no \(x^2\) term.- Use \(-1\) as the divisor because it represents \(x + 1\).Perform synthetic division:\[ \begin{array}{r|rrrr}-1 & 1 & 0 & -4 & -3 \ & & -1 & 1 & 3 \\hline & 1 & -1 & -3 & 0 \\end{array}\]The remainder is 0, confirming that \(x = -1\) is a root.

Step 3: Find Other Possible Values of x

The quotient from the synthetic division is \(x^2 - x - 3\). Solve for roots by using the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] Substituting \(a = 1\), \(b = -1\), and \(c = -3\) we get:\[x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1}\]\[x = \frac{1 \pm \sqrt{1 + 12}}{2}\]\[x = \frac{1 \pm \sqrt{13}}{2}\]Thus, the other possible values of \(x\) are \(\frac{1 + \sqrt{13}}{2}\) and \(\frac{1 - \sqrt{13}}{2}\).

Key Concepts

Synthetic Division

Synthetic division is a simple method used for dividing a polynomial by a binomial of the form \((x - c)\). It is a streamlined version of the traditional long division method, making it easier and faster for many students to grasp.

Here's how synthetic division works:

• First, identify the coefficients of the polynomial you want to divide. For instance, in our example, we divide the polynomial \(x^3 - 4x - 3\) by \((x + 1)\). The coefficients are \(1, 0, -4, -3\). Note that the \(x^2\) term is missing, which is why it's represented by 0.
• The divisor is the opposite of \(c\) in \((x - c)\). Here, since the divisor is \((x + 1)\), we use \(-1\).
• Perform the calculations: start by bringing down the leading coefficient, \(1\), then multiply it by \(-1\), and continue through the row.

The remainder must be zero if \(x = -1\) is a root, which confirms that \(x - 1\) divides the polynomial evenly.

This division gave us \(x^2 - x - 3\). Now, we can use this result in further calculations.

Polynomial Equations

Polynomial equations are algebraic equations that consist of variables raised to whole number exponents and coefficients. These are often written in a descending order based on the power of the variable, such as \(ax^n + bx^{n-1} + cx^{n-2} + \, ... \, + z = 0\).

In the context of the current exercise, we started with a polynomial equation of degree 3: \(x^3 - 4x - 3 = 0\). This equation was derived from substituting into an elliptic curve equation.

Using synthetic division, we factored the polynomial to find one of its roots, which was \(x = -1\). After factoring out \((x + 1)\), we are left with a quadratic equation \(x^2 - x - 3\).

Polynomial equations can have multiple solutions, and solving them involves factoring, synthetic division, or employing different algebraic methods suited to the nature of the variable powers.

Quadratic Formula

The quadratic formula is a powerful tool used for finding the roots of a quadratic equation, which is an equation of the form \(ax^2 + bx + c = 0\). It provides a reliable way to find solutions for quadratic equations that aren't amenable to simple factoring.

The formula is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]In this context, after applying synthetic division, we are left with the quadratic \(x^2 - x - 3 = 0\). Here, \(a = 1\), \(b = -1\), and \(c = -3\).

Substituting these values into the quadratic formula gives:
\[x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1}\]
This simplifies to:
\[x = \frac{1 \pm \sqrt{13}}{2}\]

This results in two potential solutions: \(\frac{1 + \sqrt{13}}{2}\) and \(\frac{1 - \sqrt{13}}{2}\). These solutions are the other possible values for \(x\) in the original polynomial equation.