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-√3i α + β = 4 |
α β = (2+√3i) (2-√3i) α β = 4-3(i)^{2} α β = 4+3 = 7 |
The least degree polynomial will be a quadratic polynomial.
x^{2} - (α + β)x + αβ = 0
x^{2} - 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 - 2i α + β = 6 |
α β = (3 + 2i)(3 - 2i) α β = 9 - 4i^{2} α β = 9 - 4(-1) α β = 13 |
x^{2} - 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 + √3 α + β = 2√5 |
α β = (√5 - √3)(√5 + √3) αβ = 5 - 3 αβ = 2 |
x^{2} - 2√5x + 2 = 0
α = -√5 + √3 and β = -√5 - √3
α + β = -√5 + √3 -√5 - √3 α + β = -2√5 |
α β = (-√5 + √3)(-√5 - √3) αβ = 5 - 3 αβ = 2 |
x^{2} + 2√5x + 2 = 0
The product of the quadratic polynomials.
[(x^{2} + 2) - 2√5x] [(x^{2} + 2) + 2√5x] = 0
(x^{2} + 2)^{2} - (2√5x)^{2 }= 0
x^{4} + 4 + 4x^{2} - 4(5)x^{2} = 0
x^{4} + 4 + 4x^{2} - 20x^{2} = 0
x^{4} - 16x^{2} + 4 = 0
Problem 4 :
If k is real, discuss the nature of the roots of the polynomial equation 2x^{2} + kx + k = 0 in terms of k.
Solution :
2x^{2} + kx + k = 0
Nature of roots = b^{2} - 4ac
a = 2, b = k and c = k
= k^{2} - 4(2)(k)
= k^{2} - 8k
= k(k - 8)
Problem 5 :
Prove that a straight line and parabola cannot intersect at more than two points.
Solution :
Standard form of parabola : y^{2} = 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
m^{2}x^{2} + 2mbx + b^{2} = 4ax
m^{2}x^{2} + 2mbx - 4ax + b^{2} = 0
m^{2}x^{2} + 2x(mb - 2a) + b^{2} = 0
While solving any quadratic equation, we will get two solutions.
So, the parabola and straight will intersect at two points.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM