Samacheer Kalvi Class 10 Maths Solutions Chapter 3 Algebra Exercise 3.13

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.13

Question 1. Determine the nature of the roots for the following quadratic equations
(i) \( 15x^2 + 11x + 2 = 0 \)
Answer:
To find the nature of the roots, we use the discriminant formula \( \Delta = b^2 - 4ac \). Here, \( a=15 \), \( b=11 \), and \( c=2 \). We calculate \( \Delta = (11)^2 - 4(15)(2) = 121 - 120 = 1 \). Since \( \Delta = 1 \) which is greater than 0, the equation has two distinct real roots. Knowing the discriminant helps predict how many real solutions an equation will have.
In simple words: We check a special number called the discriminant. If it is more than zero, the equation has two different real answers.

🎯 Exam Tip: Remember that a positive discriminant (\( \Delta > 0 \)) means two distinct real roots, a zero discriminant (\( \Delta = 0 \)) means two equal real roots, and a negative discriminant (\( \Delta < 0 \)) means no real roots (complex roots).

(ii) \( x^2 - x - 1 = 0 \)
Answer:
For this equation, we have \( a=1 \), \( b=-1 \), and \( c=-1 \). Calculating the discriminant: \( \Delta = (-1)^2 - 4(1)(-1) = 1 + 4 = 5 \). Since \( \Delta = 5 \) is greater than 0, the quadratic equation will have real and distinct roots. This means there are two different solutions that are real numbers.
In simple words: The discriminant for this equation is 5. Because 5 is a positive number, there are two different real answers for \( x \).

🎯 Exam Tip: Pay close attention to negative signs when substituting values into the discriminant formula, especially for \( b^2 \) and \( -4ac \).

(iii) \( \sqrt{2} t^2 - 3t + 3\sqrt{2} = 0 \)
Answer:
In this quadratic equation, \( a=\sqrt{2} \), \( b=-3 \), and \( c=3\sqrt{2} \). We compute the discriminant as \( \Delta = (-3)^2 - 4(\sqrt{2})(3\sqrt{2}) = 9 - 4(3 \times 2) = 9 - 24 = -15 \). Since \( \Delta = -15 \) is less than 0, the equation has no real roots. This means the solutions are complex numbers, not real numbers.
In simple words: The discriminant here is -15. Because it's a negative number, this equation does not have any real number answers.

🎯 Exam Tip: Remember that \( \sqrt{a} \times \sqrt{a} = a \). Apply this property carefully when multiplying terms like \( 4(\sqrt{2})(3\sqrt{2}) \).

(iv) \( 9y^2 - 6\sqrt{2}y + 2 = 0 \)
Answer:
For the equation \( 9y^2 - 6\sqrt{2}y + 2 = 0 \), we identify \( a=9 \), \( b=-6\sqrt{2} \), and \( c=2 \). We calculate the discriminant: \( \Delta = (-6\sqrt{2})^2 - 4(9)(2) = (36 \times 2) - 72 = 72 - 72 = 0 \). Since \( \Delta = 0 \), the equation has two real and equal roots. This means there is only one distinct real number solution for \( y \).
In simple words: The discriminant is zero. This means the equation has real answers, but both answers are the same.

🎯 Exam Tip: When \( \Delta = 0 \), the quadratic equation is a perfect square. This leads to identical roots.

(v) \( 9a^2b^2x^2 - 24abcdx + 16c^2d^2 = 0 \), \( a \neq 0, b \neq 0 \)
Answer:
In this equation, \( A=9a^2b^2 \), \( B=-24abcd \), and \( C=16c^2d^2 \). We find the discriminant: \( \Delta = (-24abcd)^2 - 4(9a^2b^2)(16c^2d^2) = 576a^2b^2c^2d^2 - 576a^2b^2c^2d^2 \). This simplifies to \( \Delta = 0 \). Because the discriminant is zero, the roots of the equation are real and equal. This type of equation can be factored into a perfect square.
In simple words: When we calculate the discriminant, it comes out to zero. This means the equation has answers that are real numbers, and both answers are exactly the same.

🎯 Exam Tip: Recognize that this equation is in the form \( (Ax - B)^2 = 0 \), which directly implies real and equal roots. For instance, it's \( (3abx - 4cd)^2 = 0 \).

Question 2. Find the value(s) of 'k' for which the roots of the following equations are real and equal.
(i) \( (5k - 6)x^2 + 2kx + 1 = 0 \)
Answer:
For the roots to be real and equal, the discriminant \( \Delta \) must be zero. In the equation \( (5k - 6)x^2 + 2kx + 1 = 0 \), we have \( a = (5k - 6) \), \( b = 2k \), and \( c = 1 \).
First, we set \( b^2 - 4ac = 0 \).
\( \implies (2k)^2 - 4(5k - 6)(1) = 0 \)
\( \implies 4k^2 - 20k + 24 = 0 \)
Dividing the entire equation by 4 makes it simpler:
\( \implies k^2 - 5k + 6 = 0 \)
Now, we factorize the quadratic equation to find the values of \( k \):
\( \implies (k - 3)(k - 2) = 0 \)
This means \( k - 3 = 0 \) or \( k - 2 = 0 \). Thus, the possible values for \( k \) are 3 and 2. It is important to find the value of k that makes the equation behave in a specific way.
In simple words: For the equation to have real and equal answers, a special number called the discriminant must be zero. We solve for \( k \) using this rule and find that \( k \) can be 3 or 2.

🎯 Exam Tip: Always remember the condition for real and equal roots: \( \Delta = 0 \). Make sure to simplify the resulting quadratic equation for \( k \) before solving it.

(ii) \( kx^2 + (6k + 2)x + 16 = 0 \)
Answer:
For this equation to have real and equal roots, its discriminant must be zero. Here, \( a = k \), \( b = (6k + 2) \), and \( c = 16 \).
We set \( b^2 - 4ac = 0 \).
\( \implies (6k + 2)^2 - 4(k)(16) = 0 \)
Expand the square and simplify:
\( \implies 36k^2 + 24k + 4 - 64k = 0 \)
\( \implies 36k^2 - 40k + 4 = 0 \)
Divide the entire equation by 4 to simplify:
\( \implies 9k^2 - 10k + 1 = 0 \)
Now, we factorize this quadratic equation to find the values for \( k \). We can split the middle term:
\( \implies 9k^2 - 9k - k + 1 = 0 \)
\( \implies 9k(k - 1) - 1(k - 1) = 0 \)
\( \implies (9k - 1)(k - 1) = 0 \)
This gives us two possible values for \( k \): \( k = \frac{1}{9} \) or \( k = 1 \). The discriminant must be zero for equal roots.
In simple words: We make the discriminant equal to zero. This helps us solve for \( k \), and we find that \( k \) can be either \( \frac{1}{9} \) or 1.

🎯 Exam Tip: When solving for \( k \), ensure you fully simplify the quadratic equation before factoring to avoid errors. Remember to check both factors for \( k \).

Question 3. If the roots of \( (a - b)x^2 + (b - c)x + (c - a) = 0 \) are real and equal, then prove that b, a, c are in arithmetic progression.
Answer:
Given the quadratic equation \( (a - b)x^2 + (b - c)x + (c - a) = 0 \).
For this equation, we have \( A = (a - b) \), \( B = (b - c) \), and \( C = (c - a) \).
Since the roots are real and equal, the discriminant \( \Delta \) must be 0.
Therefore, \( B^2 - 4AC = 0 \).
\( \implies (b - c)^2 - 4(a - b)(c - a) = 0 \)
Expand the terms:
\( \implies (b^2 - 2bc + c^2) - 4(ac - a^2 - bc + ab) = 0 \)
\( \implies b^2 - 2bc + c^2 - 4ac + 4a^2 + 4bc - 4ab = 0 \)
Combine like terms:
\( \implies 4a^2 + b^2 + c^2 - 4ab + 2bc - 4ac = 0 \)
This can be rearranged into a perfect square form:
\( \implies (b+c)^2 - 4a(b+c) + 4a^2 = 0 \)
This is in the form \( X^2 - 2XY + Y^2 = (X - Y)^2 \), where \( X = (b+c) \) and \( Y = 2a \).
\( \implies [(b + c) - 2a]^2 = 0 \)
Taking the square root of both sides:
\( \implies b + c - 2a = 0 \)
\( \implies b + c = 2a \)
This equation shows that \( 2a \) is the sum of \( b \) and \( c \). In an arithmetic progression (AP), the middle term is the average of its neighbors. This condition \( 2a = b + c \) means that \( b, a, c \) are in arithmetic progression. In an AP, the difference between consecutive terms is constant.
In simple words: When the roots are real and equal, the discriminant is zero. By solving this, we get \( b + c = 2a \). This is the exact condition for three numbers to be in an arithmetic progression, proving that b, a, and c are in AP.

🎯 Exam Tip: When proving a relationship, always expand and simplify the discriminant equation carefully. The identity \( (X - Y)^2 \) is often useful for simplification.

Question 4. If a, b are real then show that the roots of the equation \( (a - b)x^2 - 6(a + b)x - 9(a - b) = 0 \) are real and unequal.
Answer:
Given the quadratic equation \( (a - b)x^2 - 6(a + b)x - 9(a - b) = 0 \).
Here, \( A = (a - b) \), \( B = -6(a + b) \), and \( C = -9(a - b) \).
To determine the nature of the roots, we calculate the discriminant \( \Delta = B^2 - 4AC \).
\( \implies \Delta = (-6(a + b))^2 - 4(a - b)(-9(a - b)) \)
Expand the terms:
\( \implies \Delta = 36(a + b)^2 + 36(a - b)^2 \)
Factor out 36:
\( \implies \Delta = 36[(a + b)^2 + (a - b)^2] \)
Use the identities \( (a+b)^2 = a^2 + 2ab + b^2 \) and \( (a-b)^2 = a^2 - 2ab + b^2 \).
\( \implies \Delta = 36[(a^2 + 2ab + b^2) + (a^2 - 2ab + b^2)] \)
\( \implies \Delta = 36[2a^2 + 2b^2] \)
\( \implies \Delta = 72(a^2 + b^2) \)
Since \( a \) and \( b \) are real numbers, \( a^2 \) is always greater than or equal to 0, and \( b^2 \) is always greater than or equal to 0. Therefore, \( a^2 + b^2 \) will always be greater than or equal to 0. If \( a=0 \) and \( b=0 \), the original equation becomes \( 0 = 0 \), which is not a quadratic equation. So, for a valid quadratic, \( a \) and \( b \) cannot both be zero. Hence, \( a^2 + b^2 \) must be strictly greater than 0, which means \( \Delta = 72(a^2 + b^2) \) will always be greater than 0.
Since \( \Delta > 0 \), the roots of the equation are real and unequal. The discriminant being positive indicates two distinct real solutions for x.
In simple words: We found the discriminant \( \Delta = 72(a^2 + b^2) \). Since \( a \) and \( b \) are real, \( a^2 + b^2 \) is always positive (for a quadratic). Because \( \Delta \) is always positive, the equation will have two different real number answers.

🎯 Exam Tip: When dealing with general coefficients like \( a \) and \( b \), always simplify the discriminant completely. Pay attention to conditions that ensure the leading coefficient is non-zero and \( \Delta \) is strictly positive.

Question 5. If the roots of the equation \( (c^2 - ab)x^2 - 2(a^2 - bc)x + b^2 - ac = 0 \) are real and equal prove that either \( a = 0 \) (or) \( a^3 + b^3 + c^3 = 3abc \).
Answer:
Given the quadratic equation \( (c^2 - ab)x^2 - 2(a^2 - bc)x + b^2 - ac = 0 \).
Here, \( A = (c^2 - ab) \), \( B = -2(a^2 - bc) \), and \( C = (b^2 - ac) \).
Since the roots are real and equal, the discriminant \( \Delta \) must be 0.
Therefore, \( B^2 - 4AC = 0 \).
\( \implies [-2(a^2 - bc)]^2 - 4(c^2 - ab)(b^2 - ac) = 0 \)
Expand the terms:
\( \implies 4(a^2 - bc)^2 - 4(c^2 - ab)(b^2 - ac) = 0 \)
Divide the entire equation by 4:
\( \implies (a^2 - bc)^2 - (c^2 - ab)(b^2 - ac) = 0 \)
Expand the squared term and the product of the two binomials:
\( \implies (a^4 - 2a^2bc + b^2c^2) - (c^2b^2 - c^3a - ab^3 + a^2bc) = 0 \)
Distribute the negative sign:
\( \implies a^4 - 2a^2bc + b^2c^2 - c^2b^2 + c^3a + ab^3 - a^2bc = 0 \)
Combine like terms. The \( b^2c^2 \) terms cancel out:
\( \implies a^4 - 3a^2bc + ac^3 + ab^3 = 0 \)
Factor out \( a \) from the expression:
\( \implies a(a^3 - 3abc + c^3 + b^3) = 0 \)
Rearrange the terms inside the parenthesis:
\( \implies a(a^3 + b^3 + c^3 - 3abc) = 0 \)
This implies that either \( a = 0 \) or \( a^3 + b^3 + c^3 - 3abc = 0 \).
So, either \( a = 0 \) or \( a^3 + b^3 + c^3 = 3abc \). This proves the required condition. The condition for equal roots simplifies nicely to this algebraic identity.
In simple words: Since the roots are real and equal, the discriminant is zero. After doing all the math steps and simplifying, we get \( a(a^3 + b^3 + c^3 - 3abc) = 0 \). This means either \( a \) is zero, or the expression \( a^3 + b^3 + c^3 \) equals \( 3abc \).

🎯 Exam Tip: This question relies heavily on algebraic expansion and simplification. Be meticulous with distributing terms and combining like terms. Recognizing the cubic identity \( a^3 + b^3 + c^3 - 3abc \) is key.