Discussing the Nature of Roots in Terms of Variable :
In this section, you will learn, how to examine the nature of roots in terms of variable.
Problem 1 :
If k is real, discuss the nature of the roots of the polynomial equation 2x2 + kx + k = 0, in terms of k .
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)
Problem 2 :
Find a polynomial equation of minimum degree with rational coefficients, having 2 + √3 i as a root.
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
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.
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
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.
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
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
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.
Equation of parabola y2 = 4ax -----(1)
Equation of the line y = mx + c ------(2)
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.
After having gone through the stuff given above, we hope that the students would have understood, how to examine the nature of the roots in terms of variable.
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