Transformation of Equations and Reciprocal Equations

Transformation of Equations

Transformation of equations is a technique used to change an equation into a new one with related roots. This process is often employed to simplify an equation, to eliminate certain terms (like the $x^{n-1}$ term), or to convert it into a standard form that can be solved more easily. The key idea is to establish a relationship between the roots of the original equation and the roots of the transformed equation.

1. Equation whose roots are the reciprocals of the roots of a given equation

Let the given equation be $P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0 = 0$, with roots $\alpha_1, \alpha_2, \dots, \alpha_n$. We want to find a new equation whose roots are $\frac{1}{\alpha_1}, \frac{1}{\alpha_2}, \dots, \frac{1}{\alpha_n}$. Let $y = \frac{1}{x}$. Then $x = \frac{1}{y}$. Substituting this into the original equation: $a_n \left(\frac{1}{y}\right)^n + a_{n-1} \left(\frac{1}{y}\right)^{n-1} + \dots + a_1 \left(\frac{1}{y}\right) + a_0 = 0$ Multiply the entire equation by $y^n$ to clear the denominators: $a_n + a_{n-1} y + \dots + a_1 y^{n-1} + a_0 y^n = 0$ Rearranging the terms in descending powers of $y$: $a_0 y^n + a_1 y^{n-1} + \dots + a_{n-1} y + a_n = 0$ This is the required equation whose roots are the reciprocals of the roots of the original equation.

Example: Find the equation whose roots are the reciprocals of the roots of $x^3 - 6x^2 + 11x - 6 = 0$. The original equation has coefficients $a_3=1, a_2=-6, a_1=11, a_0=-6$. Reversing the coefficients, the new equation is $-6y^3 + 11y^2 - 6y + 1 = 0$. Multiplying by -1 to make the leading coefficient positive, we get $6y^3 - 11y^2 + 6y - 1 = 0$. The roots of $x^3 - 6x^2 + 11x - 6 = 0$ are 1, 2, 3. The reciprocals are $\frac{1}{1}=1, \frac{1}{2}, \frac{1}{3}$. Let's check if $y=1$ is a root of $6y^3 - 11y^2 + 6y - 1 = 0$: $6(1)^3 - 11(1)^2 + 6(1) - 1 = 6 - 11 + 6 - 1 = 0$. Yes. Let's check if $y=\frac{1}{2}$ is a root: $6(\frac{1}{8}) - 11(\frac{1}{4}) + 6(\frac{1}{2}) - 1 = \frac{6}{8} - \frac{11}{4} + 3 - 1 = \frac{3}{4} - \frac{11}{4} + 2 = -\frac{8}{4} + 2 = -2 + 2 = 0$. Yes. Let's check if $y=\frac{1}{3}$ is a root: $6(\frac{1}{27}) - 11(\frac{1}{9}) + 6(\frac{1}{3}) - 1 = \frac{6}{27} - \frac{11}{9} + 2 - 1 = \frac{2}{9} - \frac{11}{9} + 1 = -\frac{9}{9} + 1 = -1 + 1 = 0$. Yes.

2. Equation whose roots are $k$ times the roots of a given equation

Let the given equation be $P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0 = 0$, with roots $\alpha_1, \alpha_2, \dots, \alpha_n$. We want to find a new equation whose roots are $k\alpha_1, k\alpha_2, \dots, k\alpha_n$. Let $y = kx$. Then $x = \frac{y}{k}$. Substituting this into the original equation: $a_n \left(\frac{y}{k}\right)^n + a_{n-1} \left(\frac{y}{k}\right)^{n-1} + \dots + a_1 \left(\frac{y}{k}\right) + a_0 = 0$ Multiply the entire equation by $k^n$ to clear the denominators: $a_n y^n + a_{n-1} k y^{n-1} + \dots + a_1 k^{n-1} y + a_0 k^n = 0$ This is the required equation whose roots are $k$ times the roots of the original equation.

Example: Find the equation whose roots are 3 times the roots of $x^3 - 2x^2 + 5x - 7 = 0$. Here $k=3$. The coefficients are $a_3=1, a_2=-2, a_1=5, a_0=-7$. The new equation is $a_3 y^3 + a_2 (3) y^2 + a_1 (3^2) y + a_0 (3^3) = 0$. $1 \cdot y^3 + (-2)(3) y^2 + (5)(9) y + (-7)(27) = 0$. $y^3 - 6y^2 + 45y - 189 = 0$.

3. Equation whose roots are $m-\alpha$ for roots $\alpha$ of a given equation

Let the given equation be $P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_0 = 0$, with roots $\alpha_1, \alpha_2, \dots, \alpha_n$. We want to find a new equation whose roots are $m-\alpha_1, m-\alpha_2, \dots, m-\alpha_n$. Let $y = m-x$. Then $x = m-y$. Substituting this into the original equation: $a_n (m-y)^n + a_{n-1} (m-y)^{n-1} + \dots + a_1 (m-y) + a_0 = 0$. This equation in $y$ is the required transformed equation.

Example: Find the equation whose roots are 5 minus the roots of $x^2 - 4x + 3 = 0$. The roots of $x^2 - 4x + 3 = 0$ are 1 and 3. The new roots should be $5-1=4$ and $5-3=2$. The new equation should have roots 4 and 2. Substitute $x = 5-y$ into the equation: $(5-y)^2 - 4(5-y) + 3 = 0$ $(25 - 10y + y^2) - (20 - 4y) + 3 = 0$ $25 - 10y + y^2 - 20 + 4y + 3 = 0$ $y^2 + (-10+4)y + (25-20+3) = 0$ $y^2 - 6y + 8 = 0$. Let's check the roots of $y^2 - 6y + 8 = 0$. Factoring, we get $(y-4)(y-2)=0$, so $y=4$ and $y=2$. This matches our expected roots.

4. Equation whose roots are $\alpha - m$ for roots $\alpha$ of a given equation

Let the given equation be $P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_0 = 0$, with roots $\alpha_1, \alpha_2, \dots, \alpha_n$. We want to find a new equation whose roots are $\alpha_1-m, \alpha_2-m, \dots, \alpha_n-m$. Let $y = x-m$. Then $x = y+m$. Substituting this into the original equation: $a_n (y+m)^n + a_{n-1} (y+m)^{n-1} + \dots + a_1 (y+m) + a_0 = 0$. This equation in $y$ is the required transformed equation.

Example: Find the equation whose roots are decreased by 2, from the equation $x^3 - 9x^2 + 20x - 12 = 0$. The roots of $x^3 - 9x^2 + 20x - 12 = 0$ are 1, 2, 6. The new roots should be $1-2=-1$, $2-2=0$, $6-2=4$. Substitute $x = y+2$ into the equation: $(y+2)^3 - 9(y+2)^2 + 20(y+2) - 12 = 0$ $(y^3 + 6y^2 + 12y + 8) - 9(y^2 + 4y + 4) + (20y + 40) - 12 = 0$ $y^3 + 6y^2 + 12y + 8 - 9y^2 - 36y - 36 + 20y + 40 - 12 = 0$ $y^3 + (6-9)y^2 + (12-36+20)y + (8-36+40-12) = 0$ $y^3 - 3y^2 - 4y + 0 = 0$ $y^3 - 3y^2 - 4y = 0$. Let's check the roots of $y^3 - 3y^2 - 4y = 0$. Factoring out $y$: $y(y^2 - 3y - 4) = 0$. So, one root is $y=0$. Factoring the quadratic: $y(y-4)(y+1) = 0$. The roots are $y=0, y=4, y=-1$. This matches our expected roots.

5. Equation whose roots are $m\alpha$ for roots $\alpha$ of a given equation

This is a special case of transformation where roots are multiplied by a constant. Let the given equation be $P(x) = a_n x^n + \dots + a_0 = 0$ with roots $\alpha_1, \dots, \alpha_n$. We want roots $m\alpha_1, \dots, m\alpha_n$. Let $y = mx$, so $x = y/m$. Substituting $x=y/m$ into $P(x)=0$ and multiplying by $m^n$ gives: $a_n y^n + a_{n-1} m y^{n-1} + \dots + a_1 m^{n-1} y + a_0 m^n = 0$.

Example: Find the equation whose roots are twice the roots of $x^3 - 6x^2 + 11x - 6 = 0$. Here $m=2$. The equation is $a_3 y^3 + a_2(2)y^2 + a_1(2^2)y + a_0(2^3) = 0$. $1 \cdot y^3 + (-6)(2)y^2 + (11)(4)y + (-6)(8) = 0$. $y^3 - 12y^2 + 44y - 48 = 0$. The roots of the original equation are 1, 2, 3. The new roots should be $2 \times 1=2$, $2 \times 2=4$, $2 \times 3=6$. Let's check if $y=2$ is a root of the new equation: $2^3 - 12(2^2) + 44(2) - 48 = 8 - 12(4) + 88 - 48 = 8 - 48 + 88 - 48 = 96 - 96 = 0$. Yes. Check $y=4$: $4^3 - 12(4^2) + 44(4) - 48 = 64 - 12(16) + 176 - 48 = 64 - 192 + 176 - 48 = 240 - 240 = 0$. Yes. Check $y=6$: $6^3 - 12(6^2) + 44(6) - 48 = 216 - 12(36) + 264 - 48 = 216 - 432 + 264 - 48 = 480 - 480 = 0$. Yes.

6. Eliminating the $x^{n-1}$ term

For a general polynomial equation $a_n x^n + a_{n-1} x^{n-1} + \dots + a_0 = 0$, the sum of the roots is $-\frac{a_{n-1}}{a_n}$. To eliminate the $x^{n-1}$ term, we want the sum of the new roots to be zero. Let $y = x - h$. Then $x = y+h$. The sum of the roots $\alpha_i$ is $\sum \alpha_i = -\frac{a_{n-1}}{a_n}$. The sum of the new roots $y_i = \alpha_i - h$ is $\sum y_i = \sum (\alpha_i - h) = \sum \alpha_i - nh$. We want $\sum y_i = 0$. So, $\sum \alpha_i - nh = 0$, which means $nh = \sum \alpha_i$. Therefore, $h = \frac{1}{n} \sum \alpha_i = -\frac{a_{n-1}}{na_n}$. So, substitute $x = y - \frac{a_{n-1}}{na_n}$ (or $x = y+h$ where $h = -\frac{a_{n-1}}{na_n}$) into the original equation to get a new equation in $y$ where the $y^{n-1}$ term is absent.

Example: Transform the equation $x^3 - 6x^2 + 11x - 6 = 0$ to remove the $x^2$ term. Here $n=3$, $a_3=1$, $a_2=-6$. The value of $h$ to be subtracted is $h = -\frac{a_{n-1}}{na_n} = -\frac{-6}{3 \times 1} = \frac{6}{3} = 2$. So, we substitute $x = y+2$. $(y+2)^3 - 6(y+2)^2 + 11(y+2) - 6 = 0$ $(y^3 + 6y^2 + 12y + 8) - 6(y^2 + 4y + 4) + (11y + 22) - 6 = 0$ $y^3 + 6y^2 + 12y + 8 - 6y^2 - 24y - 24 + 11y + 22 - 6 = 0$ $y^3 + (6-6)y^2 + (12-24+11)y + (8-24+22-6) = 0$ $y^3 + 0y^2 - y + 0 = 0$ $y^3 - y = 0$. The new equation is $y^3 - y = 0$. The $y^2$ term is indeed eliminated. The roots of the original equation are 1, 2, 3. The sum is 6. The new roots should be $1-2=-1$, $2-2=0$, $3-2=1$. The roots of $y^3 - y = 0$ are $y(y^2-1)=0 \implies y(y-1)(y+1)=0$, which gives $y=0, y=1, y=-1$. This matches.

Reciprocal Equations

A reciprocal equation is a polynomial equation where if $\alpha$ is a root, then $\frac{1}{\alpha}$ is also a root. This property implies a symmetry in the coefficients of the equation.

Types of Reciprocal Equations

Reciprocal equations are classified into two types based on their coefficients:

• Type I: The coefficients are symmetric. That is, $a_k = a_{n-k}$ for all $k$. For example, $a_0=a_n, a_1=a_{n-1}, \dots$.
• Type II: The coefficients are anti-symmetric. That is, $a_k = -a_{n-k}$ for all $k$. For example, $a_0=-a_n, a_1=-a_{n-1}, \dots$.

Properties of Reciprocal Equations

1. If $P(x) = 0$ is a reciprocal equation, and $\alpha$ is a root, then $\frac{1}{\alpha}$ is also a root.
2. For Type I reciprocal equations:
   • If the degree $n$ is odd, then $(x+1)$ is a factor, meaning $x=-1$ is always a root.
   • If the degree $n$ is even, the equation can be reduced to a standard form by dividing by $x^{n/2}$ and substituting $y = x + \frac{1}{x}$.
3. For Type II reciprocal equations:
   • If the degree $n$ is odd, then $(x+1)$ is a factor, meaning $x=-1$ is always a root. The remaining equation is a Type I reciprocal equation of even degree.
   • If the degree $n$ is even, then $(x^2-1)$ is a factor, meaning $x=1$ and $x=-1$ are always roots. The remaining equation is a Type I reciprocal equation of even degree.
   • In Type II equations, the middle coefficient(s) must be zero. For example, if $n$ is even, $a_{n/2} = -a_{n-n/2} = -a_{n/2}$, so $2a_{n/2}=0 \implies a_{n/2}=0$.

Solving Reciprocal Equations

Case 1: Type I Reciprocal Equations

Odd Degree: If the degree $n$ is odd, $(x+1)$ is a factor. Divide the polynomial by $(x+1)$ to get a polynomial of degree $n-1$ (which is even). This resulting polynomial will be a Type I reciprocal equation of even degree.

Example: Solve $2x^5 + 5x^4 + 3x^3 + 3x^2 + 5x + 2 = 0$. This is a Type I reciprocal equation of odd degree (5). So, $x=-1$ is a root. Divide $2x^5 + 5x^4 + 3x^3 + 3x^2 + 5x + 2$ by $(x+1)$ using synthetic division: ``` -1 | 2 5 3 3 5 2 | -2 -3 0 -3 -2 ----------------------- 2 3 0 3 2 0 ``` The remaining equation is $2x^4 + 3x^3 + 0x^2 + 3x + 2 = 0$. This is a Type I reciprocal equation of even degree (4).

Even Degree: If the degree $n$ is even, divide the equation by $x^{n/2}$. Group terms as $(x^k + \frac{1}{x^k})$. Substitute $y = x + \frac{1}{x}$. Note that $x^2 + \frac{1}{x^2} = y^2 - 2$, $x^3 + \frac{1}{x^3} = y^3 - 3y$, and so on. This will transform the equation into a polynomial in $y$ of degree $n/2$. Solve for $y$, and then for each value of $y$, solve $x + \frac{1}{x} = y$, which is $x^2 - yx + 1 = 0$.

Continuing the example: $2x^4 + 3x^3 + 0x^2 + 3x + 2 = 0$. Divide by $x^2$ (since $n=4$, $n/2=2$): $2x^2 + 3x + 0 + \frac{3}{x} + \frac{2}{x^2} = 0$ Group terms: $2(x^2 + \frac{1}{x^2}) + 3(x + \frac{1}{x}) = 0$. Let $y = x + \frac{1}{x}$. Then $y^2 = (x + \frac{1}{x})^2 = x^2 + 2 + \frac{1}{x^2}$, so $x^2 + \frac{1}{x^2} = y^2 - 2$. Substitute into the equation: $2(y^2 - 2) + 3y = 0$ $2y^2 - 4 + 3y = 0$ $2y^2 + 3y - 4 = 0$. Solve this quadratic for $y$ using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $y = \frac{-3 \pm \sqrt{3^2 - 4(2)(-4)}}{2(2)} = \frac{-3 \pm \sqrt{9 + 32}}{4} = \frac{-3 \pm \sqrt{41}}{4}$. So, we have two values for $y$: $y_1 = \frac{-3 + \sqrt{41}}{4}$ and $y_2 = \frac{-3 - \sqrt{41}}{4}$.

Now, for each value of $y$, solve $x + \frac{1}{x} = y$, which is $x^2 - yx + 1 = 0$. For $y_1 = \frac{-3 + \sqrt{41}}{4}$: $x^2 - \left(\frac{-3 + \sqrt{41}}{4}\right)x + 1 = 0$ $4x^2 - (-3 + \sqrt{41})x + 4 = 0$. The roots are $x = \frac{-(-3 + \sqrt{41}) \pm \sqrt{(-3 + \sqrt{41})^2 - 4(4)(4)}}{2(4)} = \frac{3 - \sqrt{41} \pm \sqrt{9 - 6\sqrt{41} + 41 - 64}}{8} = \frac{3 - \sqrt{41} \pm \sqrt{-14 - 6\sqrt{41}}}{8}$. For $y_2 = \frac{-3 - \sqrt{41}}{4}$: $x^2 - \left(\frac{-3 - \sqrt{41}}{4}\right)x + 1 = 0$ $4x^2 - (-3 - \sqrt{41})x + 4 = 0$. The roots are $x = \frac{-(-3 - \sqrt{41}) \pm \sqrt{(-3 - \sqrt{41})^2 - 4(4)(4)}}{2(4)} = \frac{3 + \sqrt{41} \pm \sqrt{9 + 6\sqrt{41} + 41 - 64}}{8} = \frac{3 + \sqrt{41} \pm \sqrt{-14 + 6\sqrt{41}}}{8}$. The five roots are $-1$ (from the odd degree property), and the four roots obtained from the two values of $y$.

Case 2: Type II Reciprocal Equations

Odd Degree: If the degree $n$ is odd, $(x+1)$ is a factor. Divide by $(x+1)$. The resulting equation is a Type I reciprocal equation of even degree.

Even Degree: If the degree $n$ is even, $(x^2-1)$ is a factor, meaning $x=1$ and $x=-1$ are roots. Divide the polynomial by $(x^2-1)$. The resulting equation is a Type I reciprocal equation of degree $n-2$ (which is even).

Example: Solve $x^5 - 2x^4 - 3x^3 + 3x^2 + 2x - 1 = 0$. This is a Type II reciprocal equation of odd degree (5). So, $(x+1)$ is a factor. Using synthetic division with root -1: ``` -1 | 1 -2 -3 3 2 -1 | -1 3 0 -3 1 ----------------------- 1 -3 0 3 -1 0 ``` The remaining equation is $x^4 - 3x^3 + 0x^2 + 3x - 1 = 0$. This is a Type I reciprocal equation of even degree (4). Divide by $x^2$: $x^2 - 3x + 0 + \frac{3}{x} - \frac{1}{x^2} = 0$ Group terms: $(x^2 - \frac{1}{x^2}) - 3(x - \frac{1}{x}) = 0$. Let $z = x - \frac{1}{x}$. Then $z^2 = (x - \frac{1}{x})^2 = x^2 - 2 + \frac{1}{x^2}$, so $x^2 + \frac{1}{x^2} = z^2 + 2$. Wait, the grouping is $(x^2 - \frac{1}{x^2})$, this is not directly related to $z = x - \frac{1}{x}$ in a simple polynomial form. Let's recheck the definition of Type II. $a_k = -a_{n-k}$. For $n=4$: $a_0 = -a_4$, $a_1 = -a_3$. The equation is $a_4 x^4 + a_3 x^3 + a_2 x^2 + a_1 x + a_0 = 0$. $a_4 x^4 - a_4 x^3 + a_2 x^2 - a_4 x + a_4 = 0$. (assuming $a_4 \neq 0$) If $a_4=1$, $x^4 - x^3 + a_2 x^2 - x + 1 = 0$. In our example: $x^4 - 3x^3 + 0x^2 + 3x - 1 = 0$. Here $a_4 = 1$, $a_3 = -3$, $a_2 = 0$, $a_1 = 3$, $a_0 = -1$. Check condition: $a_0 = -1$, $a_4 = 1$. So $a_0 = -a_4$. Correct. $a_1 = 3$, $a_3 = -3$. So $a_1 = -a_3$. Correct. $a_2 = 0$. For even degree $n$, $a_{n/2}$ must be 0. Here $n=4$, $n/2=2$. $a_2=0$. Correct. So, this is indeed a Type II reciprocal equation of even degree. The property states that $(x^2-1)$ is a factor. Let's check. If $x=1$: $1 - 3 + 0 + 3 - 1 = 0$. Yes, $x=1$ is a root. If $x=-1$: $1 - 3(-1) + 0 + 3(-1) - 1 = 1 + 3 - 3 - 1 = 0$. Yes, $x=-1$ is a root. So, $(x-1)$ and $(x+1)$ are factors, meaning $(x^2-1)$ is a factor. Divide $x^4 - 3x^3 + 3x - 1$ by $(x^2-1)$. Using polynomial long division or synthetic division twice: We already know $x=-1$ is a root of the original equation, so it's a root of the reduced equation $x^4 - 3x^3 + 3x - 1 = 0$. Divide $x^4 - 3x^3 + 0x^2 + 3x - 1$ by $(x+1)$: ``` -1 | 1 -3 0 3 -1 | -1 4 -4 1 ------------------- 1 -4 4 -1 0 ``` The remaining equation is $x^3 - 4x^2 + 4x - 1 = 0$. Now divide this by $(x-1)$: ``` 1 | 1 -4 4 -1 | 1 -3 1 ---------------- 1 -3 1 0 ``` The remaining equation is $x^2 - 3x + 1 = 0$. This is a Type I reciprocal equation of even degree (2). Solve $x^2 - 3x + 1 = 0$: $x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(1)}}{2(1)} = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2}$. The roots of the original equation $x^5 - 2x^4 - 3x^3 + 3x^2 + 2x - 1 = 0$ are: $-1$ (from the odd degree property), $1$ (from the $x^2-1$ factor), $-1$ (repeated root from the $x^2-1$ factor), $\frac{3 + \sqrt{5}}{2}$, $\frac{3 - \sqrt{5}}{2}$. So the roots are $-1$ (with multiplicity 2), $1$, $\frac{3 + \sqrt{5}}{2}$, $\frac{3 - \sqrt{5}}{2}$.