**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 2x^{2} + kx + k = 0, in terms of k .

**Solution :**

To find the nature of the roots, we have to use the formula for discriminant

Δ = b^{2} - 4ac

a = 2, b = k and c = k

Δ = k^{2} - 4(2) (k)

Δ = k^{2} - 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)

= 2^{2} - 3i^{2}

= 4 + 3

= 7

Thus x^{2} − 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 - 4i^{2}

= 9 + 4

= 13

Hence, the required polynomial is x^{2} − 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 x^{2} - 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 x^{2} + 2√5 + 2 = 0

Multiplying both quadratic equations, we get

= (x^{2} + 2 - 2√5x)(x^{2} + 2 + 2√5x)

= (x^{2} + 2)^{2} - (2√5x)^{2}

= x^{4} + 4x^{2} + 4 - 4(5)x^{2}

= x^{4} - 16x^{2} + 4

**Problem 5 :**

Prove that a straight line and parabola cannot intersect at more than two points.

**Solution :**

Equation of parabola y^{2} = 4ax -----(1)

Equation of the line y = mx + c ------(2)

From (1),

x = y^{2}/4a

By applying the value of x in (2), we get

y = m(y^{2}/4a) + c

4ay = my^{2} + 4ac

my^{2} - 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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