PROBLEMS ON NATURE OF ROOTS

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)

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 :

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.

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