# DISCUSSING THE NATURE OF ROOTS IN TERMS OF VARIABLE

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.

## Discussing the Nature of Roots in Terms of Variable - Practice Problems with Solutions

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)

## Real and unequal Roots

If Δ = 0, then the roots are real and equal.

If k > 8, then we get positive values for discriminant.

## Real and Equal Roots

If Δ = 0, then the roots are real and equal.

If k = 0 or k = 8, then the value of discriminant will become 0.

## Imaginary Roots

If Δ < 0, then the roots are imaginary.

If 0 < k < 8, then we get negative values for discriminant. Hence we have imaginary roots.

## Real 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  =  25

Product of roots  =  (5 + √3) (5 − √3)

=   5 - 3

=  2

Quadratic equation x225x + 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  =  -25

Product of roots  =  (-5 + √3) (-5 − √3)

=   5 - 3

=  2

Quadratic equation x2 + 25 + 2  =  0

Multiplying both quadratic equations, we get

=  (x2 + 2 25x)(x2 + 2 + 25x)

=  (x2 + 2)2 - (25x)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.  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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