Problem 1 :
If k is real, discuss the nature of the roots of the polynomial equation 2x2 + kx + k = 0, in terms of k .
Solution :
To find the nature of the roots, we have to use the formula for discriminant
Δ = b2 - 4ac
a = 2, b = k and c = k
Δ = k2 - 4(2) (k)
Δ = k2 - 8k
Δ = k(k - 8)
If Δ = 0, then the roots are real and equal.
If k > 8, then we get positive values for discriminant.
If Δ = 0, then the roots are real and equal.
If k = 0 or k = 8, then the value of discriminant will become 0.
If Δ < 0, then the roots are imaginary.
If 0 < k < 8, then we get negative values for discriminant. Hence we have imaginary roots.
If Δ > 0, then the roots are real.
If k < 0, then we get positive values for discriminant. Hence we have real roots.
Problem 2 :
Find a polynomial equation of minimum degree with rational coefficients, having 2 + √3 i as a root.
Solution :
Since 2 + √3i is a root of the required polynomial equation with real coefficients, 2 - √3i is also a root.
Sum of roots = 2 + √3i + 2 - √3i = 4
Product of roots = (2 + √3i) (2 - √3i)
= 22 - 3i2
= 4 + 3
= 7
Thus x2 − 4x + 7 = 0 is the required polynomial equation.
Problem 3 :
Find a polynomial equation of minimum degree with rational coefficients, having 2i + 3 as a root.
Solution :
Since the given root is complex number, its conjugate must be the another root.
The roots are 3 + 2i, 3 - 2i
Sum of roots = 3 + 2i + 3 - 2i = 6
Product of roots = (3 + 2i) (3 - 2i)
= 9 - 4i2
= 9 + 4
= 13
Hence, the required polynomial is x2 − 6x + 13 = 0.
Problem 4 :
Find a polynomial equation of minimum degree with rational coefficients, having √5 − √3 as a root.
Solution :
Note :
If √p + √q is a root of a polynomial equation with rational coefficients, then √p − √q,− √p + √q , and − √p − √q are also roots of the same polynomial equation.
Four roots of the polynomial are √5 + √3, √5 − √3, -√5 + √3, -√5 − √3
Let us find two quadratic equations by grouping the pair of roots.
√5 + √3, √5 − √3
Sum of roots = √5 + √3 + √5 − √3 = 2√5
Product of roots = (√5 + √3) (√5 − √3)
= 5 - 3
= 2
Quadratic equation x2 - 2√5x + 2 = 0
Now let us find the other quadratic equation by grouping the other two roots.
-√5 + √3, -√5 − √3
Sum of roots = -√5 + √3 - √5 − √3 = -2√5
Product of roots = (-√5 + √3) (-√5 − √3)
= 5 - 3
= 2
Quadratic equation x2 + 2√5 + 2 = 0
Multiplying both quadratic equations, we get
= (x2 + 2 - 2√5x)(x2 + 2 + 2√5x)
= (x2 + 2)2 - (2√5x)2
= x4 + 4x2 + 4 - 4(5)x2
= x4 - 16x2 + 4
Problem 5 :
Prove that a straight line and parabola cannot intersect at more than two points.
Solution :
Equation of parabola y2 = 4ax -----(1)
Equation of the line y = mx + c ------(2)
From (1),
x = y2/4a
By applying the value of x in (2), we get
y = m(y2/4a) + c
4ay = my2 + 4ac
my2 - 4ay + 4ac = 0
By solving the above equation, we get two different values of y.
Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
Apr 23, 24 09:10 PM
Apr 23, 24 12:32 PM
Apr 23, 24 12:07 PM