Samacheer Kalvi Class 11 Maths Solutions Chapter 2 Basic Algebra Exercise 2.6

Question 1. Find the zeros of the polynomial function f(x) = 4x² – 25
Answer: To find the zeros of the polynomial function \( f(x) = 4x^2 - 25 \), we need to set \( f(x) \) equal to zero.
\( 4x^2 - 25 = 0 \)
Now, we isolate the \( x^2 \) term.
\( \implies 4x^2 = 25 \)
\( \implies x^2 = \frac{25}{4} \)
To find \( x \), we take the square root of both sides, remembering to include both positive and negative solutions.
\( \implies x = \pm\sqrt{\frac{25}{4}} \)
\( \implies x = \pm\frac{5}{2} \)
Thus, the two zeros of the polynomial function are \( -\frac{5}{2} \) and \( \frac{5}{2} \).
In simple words: To find where a function is zero, we make the function equal to zero and solve for x. For this problem, we find that x can be either positive or negative five-halves. These values are the points where the graph of the function crosses the x-axis.

🎯 Exam Tip: When solving for \( x^2 = k \), always remember to include both positive and negative square roots (\( x = \pm\sqrt{k} \)) to find all possible solutions.

Question 2. If x = - 2 is one root of x³ – x² – 17x = 22, then find the other roots of equation.
Answer: First, rewrite the equation as a polynomial \( f(x) = x^3 - x^2 - 17x - 22 = 0 \).
Since \( x = -2 \) is a root, it means that \( (x + 2) \) is a factor of the polynomial. We can use synthetic division to divide the polynomial by \( (x + 2) \) and find the other factors.
Using synthetic division with the root -2 and coefficients (1, -1, -17, -22):
The coefficients of the resulting quadratic factor are 1, -3, -11, with a remainder of 0. This gives us the quadratic equation \( x^2 - 3x - 11 = 0 \).
Now, we solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
Here, \( a = 1 \), \( b = -3 \), and \( c = -11 \).
Substitute these values into the formula:
\( x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-11)}}{2(1)} \)
\( \implies x = \frac{3 \pm \sqrt{9 + 44}}{2} \)
\( \implies x = \frac{3 \pm \sqrt{53}}{2} \)
So, the other two roots of the equation are \( \frac{3 + \sqrt{53}}{2} \) and \( \frac{3 - \sqrt{53}}{2} \). A cubic equation will always have three roots, and we have found all of them.
In simple words: We are given one answer (root) for the problem. We use this answer to simplify the big math problem into a smaller one. Then, we use a special formula to find the remaining two answers. These answers are numbers that make the whole equation true when plugged in for x.

🎯 Exam Tip: When using synthetic division, ensure all powers of x are represented with coefficients, using 0 for any missing terms, to avoid errors in the division process.

Question 3. Find the real roots of x⁴ = 16.
Answer: To find the real roots of \( x^4 = 16 \), we first move all terms to one side to set the equation to zero.
\( x^4 = 16 \)
\( \implies x^4 - 16 = 0 \)
This equation can be factored as a difference of squares: \( (a^2 - b^2) = (a - b)(a + b) \). Here, \( a = x^2 \) and \( b = 4 \).
\( \implies (x^2 + 4)(x^2 - 4) = 0 \)
Now, we solve each factor for \( x \).
For the first factor, \( x^2 + 4 = 0 \):
\( x^2 = -4 \)
This equation has no real solutions because the square of any real number cannot be negative. The solutions here would be imaginary.
For the second factor, \( x^2 - 4 = 0 \):
\( x^2 = 4 \)
\( \implies x = \pm\sqrt{4} \)
\( \implies x = \pm 2 \)
Thus, the real roots of the equation \( x^4 = 16 \) are \( x = 2 \) and \( x = -2 \). These are the only numbers that when raised to the fourth power equal 16.
In simple words: We want to find the numbers that, when multiplied by themselves four times, give 16. We break the problem into two smaller parts. One part gives answers that are not "real" numbers, but the other part gives two real answers: 2 and -2.

🎯 Exam Tip: Always check if a polynomial can be factored as a difference of squares or a sum/difference of cubes, as this often simplifies finding roots, especially for higher powers.

Question 4. Solve (2x + 1)² – (3x + 2)² = 0
Answer: The given equation is \( (2x + 1)^2 - (3x + 2)^2 = 0 \).
This equation is in the form of \( a^2 - b^2 = 0 \), which can be factored as \( (a + b)(a - b) = 0 \).
Here, let \( a = (2x + 1) \) and \( b = (3x + 2) \).
Applying the formula, we get:
\( [(2x + 1) + (3x + 2)] \cdot [(2x + 1) - (3x + 2)] = 0 \)
First, simplify the terms inside the square brackets:
For the first bracket: \( (2x + 1 + 3x + 2) = (5x + 3) \)
For the second bracket: \( (2x + 1 - 3x - 2) = (-x - 1) \)
Now, the equation becomes:
\( (5x + 3)(-x - 1) = 0 \)
For the product of two factors to be zero, at least one of the factors must be zero. So, we set each factor equal to zero and solve for \( x \).
Case 1: \( 5x + 3 = 0 \)
\( \implies 5x = -3 \)
\( \implies x = -\frac{3}{5} \)
Case 2: \( -x - 1 = 0 \)
\( \implies -x = 1 \)
\( \implies x = -1 \)
Therefore, the solution set for the equation is \( \left\{ -1, -\frac{3}{5} \right\} \). This method often saves time compared to expanding each squared term and then solving the resulting quadratic equation.
In simple words: This problem looks like a common math trick called "difference of squares." We break it into two simpler parts, where each part equals zero. Solving these two simpler parts gives us the two answers for x.

🎯 Exam Tip: Always look for opportunities to apply algebraic identities like the difference of squares (\( a^2 - b^2 \)) to simplify complex equations and solve them more efficiently.