**Transformation of Equation**

**Transformation equation 1:** Transformation of an equation into another equation whose roots are the reciprocals of the roots of the given equation.

Let f(x) = aₒ xⁿ + a₁ x ⁿˉ¹ + a₂ x ⁿˉ² +. . . + a_{n} … (1)

Be the given equation. Let x and y be respectively the roots of the give equation and that of the transformed equation. Then y = 1/x

⇒ x = 1/y.

Putting x = 1/y in (i), we get

\(\frac{{{a}_{0}}}{{{y}^{n}}}+\frac{{{a}_{1}}}{{{y}^{n-1}}}+\frac{{{a}_{2}}}{{{y}^{n-2}}}+….+\frac{{{a}_{n-1}}}{y}+{{a}_{n}}=0\)

This is the required equation.

Thus, to obtain an equation whose roots are reciprocals of the roots of a given equation is obtained by replacing x by 1/x in the given equation.

**Transformation equation 2:** Transformation of an equation to another equation whose roots are negative of the roots of a given equation.

Let the given equation be f(x) = aₒ xⁿ + a₁ xⁿˉ¹ +a₂ x ⁿˉ² + . . . + a_{n }= 0

Let x be a root of the given equation and y be a root of the transformed equation. Then y = -x or x = – y. Thus, the transformed equation is obtained by putting x = -y in f(x) = 0 and is therefore f(-y) = 0

Or aₒ yⁿ – a₁ yⁿˉ¹ + . . . + (-1)ⁿa_{n}

**Example:** The equation whose roots are negative of the roots of the equation x³ -5x² -7x -3 = 0

**Solution: **f(x) = 0 then negative roots f(-x) = 0.

(-x)³ – 5(-x)² – 7(-x) – 3 = 0

x³ – 5x² + 7x – 3 = 0

x³ +5x² – 7x + 3 = 0

**Transformation equation 3:** Transformation of an equation to another equation whose roots are square of the roots of a given equation.

Let x be a root of the given equation and y be that of the transformed equation. Then, y = x²

⇒ x = √y

Thus, an equation whose roots are squares of the roots of a given equation is obtained by replacing x by √x in the given equation.

**Transformation equation 4:** Transformation of an equation to another equation whose roots are cubes of the roots of a given equation.

Let x be a root the given equation and y be that of the transformation equation. Then, y = x³

⇒ x = y^{⅓}

Thus, an equation whose roots are cubes of the roots given equation is obtained by replacing x by x⅓ in the given equation.

**Example:** Find the polynomial equation whose roots are the reciprocal of the x⁵ + 11x⁴ + x³ + 4x² – 13x + 6 = 0

**Solution: **Given that f(x) = x⁵ + 11x⁴ + x³ + 4x² -13x + 6 = 0

Required equation is f(1/x) = 0

\(\frac{1}{{{x}^{5}}}+\frac{11}{{{x}^{4}}}+\frac{1}{{{x}^{3}}}+\frac{4}{{{x}^{2}}}-\frac{13}{x}+6=0\)

Take L.C.M

1 + 11 x + x² + 4x³ – 13x⁴ + 6x⁵ = 0.