FIND POLYNOMIAL OF LEAST DEGREE WITH GIVEN ZEROS

Problem 1 :

Find a polynomial equation of minimum degree with rational coefficients having 2+√3i as a root.

Solution :

Since 2+√3i is one root of the polynomial, other root be 2-√3i

α = 2+√3i and β = 2-√3i

The least degree polynomial will be a quadratic polynomial.

x² - (α + β)x + αβ = 0

x² - 4x + 7 = 0

Problem 2 :

Find a polynomial equation of minimum degree with rational coefficients having 2i+3 as a root.

Solution :

Since 3+2i is one root of the polynomial, other root be 3-2i

α = 3+2i and β = 3-2i

x² - 6x + 13 = 0

Problem 3 :

Find a polynomial equation of minimum degree with rational coefficients having √5 - √3 as a root.

Solution :

The roots of the polynomial are √5 - √3, √5 + √3, -√5 + √3 and -√5 - √3

α = √5 - √3 and β = √5 + √3

x² - 2√5x + 2 = 0

α = -√5 + √3 and β = -√5 - √3

x² + 2√5x + 2 = 0

The product of the quadratic polynomials.

[(x² + 2) - 2√5x]  [(x² + 2) + 2√5x] = 0

(x² + 2)² - (2√5x)² = 0

x⁴ + 4 + 4x² - 4(5)x² = 0

x⁴ + 4 + 4x² - 20x² = 0

x⁴ - 16x² + 4 = 0

Problem 4 :

If k is real, discuss the nature of the roots of the polynomial equation 2x² + kx + k = 0 in terms of k.

Solution :

2x² + kx + k = 0

Nature of roots = b² - 4ac

a = 2, b = k and c = k

= k² - 4(2)(k)

= k² - 8k

= k(k - 8)

• If k < 0, then b² - 4ac > 0. So, the roots are real.
• If 0 < k < 8, then b² - 4ac < 0. So, the roots are imaginary.
• If k = 0 or k = 8, then the roots are real and equal.

Problem 5 :

Prove that a straight line and parabola cannot intersect at more than two points.

Solution :

Standard form of parabola : y² = 4ax ---(1)

Standard form of straight line : y = mx + b ----(2)

To find the point of intersection, we will solve these two equations.

(mx + b)² = 4ax

m²x² + 2mbx + b² = 4ax

m²x² + 2mbx - 4ax + b² = 0

m²x² + 2x(mb - 2a) + b² = 0

While solving any quadratic equation, we will get two solutions.

So, the parabola and straight will intersect at two points.