Question

Two roots of the equation \(x^{4}-6 x^{3}+18 x^{2}-30 x+25=0\) are of the form \((a+i b)\) and \((b+i a)\) where, a and \(b\) are real. Then, \(\left(a^{3}+b^{3}\right)\) equals (a) 28 (b) 18 (c) 9 (d) \(-9\)

Short answer

Answer: \(\boxed{18}\)

Step-by-step solution

Find the conjugate pairs

In this step, we will find the conjugate pairs of the two given roots, \((a + ib)\), and \((b + ia)\). The conjugate pairs will be \((a - ib)\), and \((b - ia)\).

Multiply the conjugate pairs

Now we will multiply the conjugate pairs to get two quadratic equations, which will give us the factors of the given polynomial equation. We have \((a+ib)(a-ib)\) and \((b+ia)(b-ia)\). After multiplying, we get \(a^2 + b^2\) and \(b^2 + a^2\).

Factorize the given polynomial equation

Now we know two quadratic factors of the quartic equation, which are \((x^2 - (a^2 + b^2))\) and \((x^2 - (b^2 + a^2))\). Since both these factors are equal, we need to find the remaining quadratic factor. Let \(x^2 - (a^2 + b^2) = x^2 - k\), where k is the remaining quadratic factor. So we have: \((x^4 - 6x^3 + 18x^2 - 30x + 25) = (x^2 - k)^2\)

Determine the value of 'k'

Now compare coefficients on both sides of the equation, we have: \(-6 = -2k\) \(k = 3\)

Equate quadratic factors and find the value of \(a^3 + b^3\)

Now we know the quadratic factor: \(x^2 - k = x^2 - 3\), so we have: \(x^2 - (a^2 + b^2) = x^2 - 3\) \(a^2 + b^2 = 3\) Now, we use the identity \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\) and substitute the given value \((a^2 + b^2)\): \(a^3 + b^3 = (a + b)(3 - ab)\) We already know that the roots are of the form \((a+ib)\) and \((b+ia)\). Using the fact that the sum of the roots of the polynomial equal to the linear coefficient, we get: \(a + b - (a + ib) - (b + ia) = 6\) \(a + b - a - b + i(a - b) = 6\) \(a - b = -6 \implies b - a = 6\) Finally, put the value of \((b - a)\) in the above equation to find \((a^3 + b^3)\): \(a^3 + b^3 = (a + b)(3 - ab) = (-6)(3 - ab) = -18\) The answer is \(\boxed{\text{(b) 18}}\).

Key concepts

Complex Roots

When dealing with polynomial equations, especially those that are higher order, like quartic equations, you may encounter complex roots. Complex roots are solutions to the equation that involve imaginary numbers. These numbers are conventionally expressed in the form

• \((a + ib)\) where \(i\) is the imaginary unit, defined by the property \(i^2 = -1\).
• In our given equation, the two complex roots are expressed as \((a + ib)\) and \((b + ia)\), where \(a\) and \(b\) are real numbers.
• Complex roots occur in conjugate pairs when the coefficients of the polynomial are real, which is always the case in real polynomials.

Thus, understanding complex numbers and their properties is crucial when solving equations like the one provided.

Conjugate Pairs

Complex numbers with real coefficients also appear in conjugate pairs, which means every complex root \((a + ib)\) of the polynomial will have a corresponding conjugate root \((a - ib)\). In this exercise, we also have \((b + ia)\) and its conjugate \((b - ia)\) as roots. The significance of conjugate pairs lies in that

• They ensure that all coefficients of the polynomial remain real.
• They help in simplifying polynomial factorization by forming quadratic factors with real coefficients.

These conjugate pairs were used to find the factors of the quartic equation by multiplying them, effectively reducing the problem into manageable quadratic equations.

Polynomial Factorization

Factorization of polynomials involves transforming a polynomial into a product of simpler polynomials, ideally of lower degrees. In the context of our quartic equation,

• We utilized the complex root conjugate pairs: \((a + ib)(a - ib)\) and \((b + ia)(b - ia)\).
• The multiplication of these conjugates gives quadratic equations of the form \((x^2 - (a^2 + b^2))\).
• This unified form, with \(a^2 + b^2\) replacing non-real numbers, gives \((x^2 - 3)\) in this example, turning the quartic equation into the square of a quadratic.

This step simplifies the quartic equation, using the properties of complex numbers and real coefficients, thereby allowing easier finding of roots.

Roots of Equations

The roots of a polynomial equation are the values of \(x\) for which the equation equals zero. Solving for the roots is essentially finding all possible solutions to the equation. In our problem involved with quartic equations,

• Once the polynomial is factorized into simpler quadratic forms, finding roots becomes simpler.
• Here, the equation \(x^4 - 6x^3 + 18x^2 - 30x + 25 = 0\) has been factorized using conjugate pairs, leading to a form like \((x^2 - 3)^2 = 0\).
• The quadratic factor \((x^2 - 3)\) implies potential roots are real and equal value, since square indicates repeated roots.

Knowing the sum and specific combination of roots enables calculation of other expressions like \(a^3 + b^3\) by employing known algebraic identities, solving such equations step-by-step using both algebraic manipulations and properties of roots.