Samacheer Kalvi Class 12 Maths Solutions Chapter 3 Theory of Equations Exercise 3.2

Question 1. If k is real, discuss the nature of the roots of the polynomial equation \(2x^2 + kx + k = 0\), in terms of k.
Answer: The given quadratic equation is \(2x^2 + kx + k = 0\).
We identify the coefficients as \(a = 2\), \(b = k\), and \(c = k\).
The discriminant, \(\Delta\), helps us determine the nature of the roots. It is calculated as \(b^2 - 4ac\).
So, \( \Delta = k^2 - 4(2)(k) \)
\( \Delta = k^2 - 8k \)

(i) If the roots are equal, the discriminant must be zero.
\( k^2 - 8k = 0 \)
\( \implies k(k - 8) = 0 \)
\( \implies k = 0 \) or \( k = 8 \)

(ii) If the roots are real, the discriminant must be greater than zero.
\( k^2 - 8k > 0 \)
\( k(k - 8) > 0 \)
This inequality holds true when \( k \) is less than 0 or \( k \) is greater than 8.
So, \( k \in (-\infty, 0) \cup (8, \infty) \)

(iii) If the roots are imaginary (not real), the discriminant must be less than zero.
\( k^2 - 8k < 0 \)
\( \implies k(k - 8) < 0 \)
This inequality holds true when \( k \) is between 0 and 8.
So, \( k \in (0, 8) \)
The discriminant helps us understand if roots are real, equal, or complex without solving the equation fully.
In simple words: We check a special number called the discriminant. If this number is zero, the answers (roots) are the same. If it's more than zero, the answers are real and different. If it's less than zero, the answers are imaginary. We find ranges for 'k' based on these conditions.

🎯 Exam Tip: Remember the three conditions for the discriminant (\(\Delta\)): \(\Delta = 0\) for equal roots, \(\Delta > 0\) for real and distinct roots, and \(\Delta < 0\) for imaginary roots. Clearly state the range of 'k' for each case.

Question 2. Find a polynomial equation of minimum degree with rational coefficients, having \(2 + \sqrt{3} i\) as a root.
Answer: If \(2 + \sqrt{3} i\) is a root, and the polynomial has rational coefficients, then its conjugate must also be a root. Complex roots always come in pairs (conjugates) when coefficients are real.
So, let the first root be \(r_1 = 2 + i\sqrt{3}\).
The other root will be its conjugate, \(r_2 = 2 - i\sqrt{3}\).

Now, we calculate the sum of the roots (S.O.R.):
S.O.R. \( = r_1 + r_2 = (2 + i\sqrt{3}) + (2 - i\sqrt{3}) = 2 + 2 = 4 \)

Next, we calculate the product of the roots (P.O.R.):
P.O.R. \( = r_1 \times r_2 = (2 + i\sqrt{3})(2 - i\sqrt{3}) \)
This is in the form \((a+b)(a-b) = a^2 - b^2\).
\( = 2^2 - (i\sqrt{3})^2 = 4 - (i^2 \times 3) = 4 - (-1 \times 3) = 4 + 3 = 7 \)

The general form of a quadratic equation is \(x^2 - (\text{S.O.R.})x + (\text{P.O.R.}) = 0\).
Substituting the values we found:
\( x^2 - 4x + 7 = 0 \)
This is the polynomial equation of minimum degree with rational coefficients.
In simple words: If a complex number like \(2 + i\sqrt{3}\) is an answer, its "partner" \(2 - i\sqrt{3}\) must also be an answer for the final equation to have normal numbers (rational coefficients). We add these two roots to get the sum, and multiply them to get the product. Then, we put these into the standard quadratic equation form to find the answer.

🎯 Exam Tip: Always remember that if a polynomial with real (or rational) coefficients has a complex root, its conjugate must also be a root. This ensures the coefficients of the polynomial remain rational.

Question 3. Find a polynomial equation of minimum degree with rational coefficients, having \(2i + 3\) as a root.
Answer: The given root is \(3 + 2i\). Since the polynomial needs to have rational coefficients, any complex roots must appear in conjugate pairs. This property ensures the polynomial itself has real coefficients.
So, if \(r_1 = 3 + 2i\) is a root, then the other root must be its conjugate, \(r_2 = 3 - 2i\).

First, calculate the sum of the roots (S.O.R.):
S.O.R. \( = r_1 + r_2 = (3 + 2i) + (3 - 2i) = 3 + 3 = 6 \)

Next, calculate the product of the roots (P.O.R.):
P.O.R. \( = r_1 \times r_2 = (3 + 2i)(3 - 2i) \)
This is in the form \((a+b)(a-b) = a^2 - b^2\).
\( = 3^2 - (2i)^2 = 9 - (4i^2) = 9 - (4 \times -1) = 9 + 4 = 13 \)

The general form of a quadratic equation is \(x^2 - (\text{S.O.R.})x + (\text{P.O.R.}) = 0\).
Substituting the values we found:
\( x^2 - (6)x + (13) = 0 \)
\( \implies x^2 - 6x + 13 = 0 \)
This is the polynomial equation of minimum degree with rational coefficients.
In simple words: When a complex number is a root, its conjugate must also be a root for the polynomial to have basic, rational numbers in its equation. We find the sum and product of these two roots and then use them to build the simplest polynomial equation.

🎯 Exam Tip: Always list the conjugate pair explicitly to show your understanding. Carefully calculate the sum and product of roots, especially when dealing with imaginary numbers, to avoid sign errors.

Question 4. Find a polynomial equation of minimum degree with rational coefficients, having \(\sqrt{5} - \sqrt{3}\) as a root.
Answer: For a polynomial to have rational coefficients, all irrational and complex roots must appear in conjugate pairs. If \(\sqrt{5} - \sqrt{3}\) is a root, then its conjugates must also be roots.
The root given is \(r_1 = \sqrt{5} - \sqrt{3}\).
Its conjugates are: \(r_2 = \sqrt{5} + \sqrt{3}\), \(r_3 = -\sqrt{5} - \sqrt{3}\), and \(r_4 = -\sqrt{5} + \sqrt{3}\).
To find an equation with rational coefficients, we can first pair \(r_1\) and \(r_2\):
Let's consider the roots \( \sqrt{5} - \sqrt{3} \) and \( \sqrt{5} + \sqrt{3} \).
Sum of these roots \( = (\sqrt{5} - \sqrt{3}) + (\sqrt{5} + \sqrt{3}) = 2\sqrt{5} \)
Product of these roots \( = (\sqrt{5} - \sqrt{3})(\sqrt{5} + \sqrt{3}) = \sqrt{5}^2 - \sqrt{3}^2 = 5 - 3 = 2 \)
This gives the quadratic factor: \( x^2 - (2\sqrt{5})x + 2 = 0 \). This equation has an irrational coefficient (\(2\sqrt{5}\)).

To make the coefficients rational, we need to eliminate the irrational term. We can achieve this by multiplying this equation by another factor derived from its conjugate form.
Consider the product of the terms \((x^2 + 2\sqrt{5}x + 2)\) and \((x^2 - 2\sqrt{5}x + 2)\).
This resembles the form \((A + B)(A - B) = A^2 - B^2\), where \(A = (x^2 + 2)\) and \(B = (2\sqrt{5}x)\).
So, the product is \( (x^2 + 2)^2 - (2\sqrt{5}x)^2 = 0 \)
\( (x^4 + 4x^2 + 4) - (4 \times 5x^2) = 0 \)
\( x^4 + 4x^2 + 4 - 20x^2 = 0 \)
\( \implies x^4 - 16x^2 + 4 = 0 \)
This is the polynomial equation of minimum degree with rational coefficients. All irrational roots must appear in conjugate pairs for the coefficients to be rational.
In simple words: When a root has square roots, its "partners" (conjugates) must also be roots to make the final equation have only whole numbers or fractions as coefficients. We first find an equation using one pair of conjugates, which still has a square root in it. Then, we multiply that equation by another form that helps remove the square root, giving us a final equation with only rational numbers.

🎯 Exam Tip: For irrational roots involving square roots, remember that if \(a + \sqrt{b}\) is a root, then \(a - \sqrt{b}\) must also be a root. If there are multiple square roots (like \(\sqrt{5} - \sqrt{3}\)), you need to consider all permutations of signs to ensure all irrationalities are removed and the final polynomial has rational coefficients.

Question 5. Prove that a straight line and parabola cannot intersect at more than two points.
Answer: Let's use the standard equations for a parabola and a straight line to see how many times they can cross.
The standard equation of a parabola is given by: \( y^2 = 4ax \) .....(1)
The equation of a straight line is given by: \( y = mx + c \) .....(2)

To find the points where they intersect, we substitute the value of \( y \) from the line equation into the parabola equation:
\( (mx + c)^2 = 4ax \)

Now, expand the left side:
\( m^2x^2 + 2mcx + c^2 = 4ax \)

Rearrange the terms to form a quadratic equation in terms of \( x \):
\( m^2x^2 + 2mcx - 4ax + c^2 = 0 \)
\( \implies m^2x^2 + (2mc - 4a)x + c^2 = 0 \)
\( \implies m^2x^2 + 2(mc - 2a)x + c^2 = 0 \)

This is a quadratic equation in \( x \). A quadratic equation can have at most two solutions for \( x \). Each solution for \( x \) corresponds to a unique point of intersection (and a corresponding \( y \) value). Since the combined equation is quadratic in x, it can only have two x-values, meaning at most two intersection points. Therefore, a straight line and a parabola cannot intersect at more than two points.
These intersection points can be real or imaginary, corresponding to visible intersections or no intersection at all.
In simple words: We put the equation of the line into the equation of the parabola. This creates a new equation which is a quadratic equation (like \(ax^2 + bx + c = 0\)). A quadratic equation can only have two solutions. So, a line and a parabola can meet at most two times.

🎯 Exam Tip: The key to this proof is recognizing that the resulting combined equation is a quadratic. Clearly state that a quadratic equation has a maximum of two roots, which directly translates to a maximum of two intersection points.