# FIND POLYNOMIAL OF LEAST DEGREE WITH GIVEN ZEROS

Let p and q be rational numbers so that √p and √q are irrational numbers; further let one of √p and √q be not a rational multiple of the other.

If √p + √q is a root of a polynomial equation with rational coefficients, then

√p - √q, -√p + √q and -√p - √q

are also roots.

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

 α + β = 2+√3i + 2-√3iα + β = 4 α β = (2+√3i) (2-√3i)α β = 4-3(i)2α β = 4+3= 7

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

 α + β = 3 + 2i + 3 - 2iα + β = 6 α β = (3 + 2i)(3 - 2i)α β = 9 - 4i2α β = 9 - 4(-1)α β = 13

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

 α + β = √5 - √3 + √5 + √3α + β = 2√5 α β = (√5 - √3)(√5 + √3)αβ = 5 - 3αβ = 2

x2 - 2√5x + 2 = 0

α = -√5 + √3 and β = -√5 - √3

 α + β = -√5 + √3 -√5 - √3α + β = -2√5 α β = (-√5 + √3)(-√5 - √3)αβ = 5 - 3αβ = 2

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.

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