RBSE Solutions Class 10 Maths Chapter 3 Polynomials Exercise 3.5

Question 1. Find the nature of roots of following quadratic equations.
(i) \( 2x^2 - 3x + 5 = 0 \)
(ii) \( 2x^2 - 4x + 3 = 0 \)
(iii) \( 2x^2 + x - 1 = 0 \)
(iv) \( x^2 - 4x + 4 = 0 \)
(v) \( 2x^2 + 5x + 5 = 0 \)
(vi) \( 3x^2 - 2x + \frac{1}{3} = 0 \)
Answer:
(i) Given equation: \( 2x^2 - 3x + 5 = 0 \)
Comparing this with the general quadratic equation \( ax^2 + bx + c = 0 \), we get:
\( a = 2, b = -3, c = 5 \)
The discriminant \( (D) \) is calculated as: \( D = b^2 - 4ac \)
\( = (-3)^2 - 4 \times 2 \times 5 \)
\( = 9 - 40 \)
\( = -31 \)
Since \( D = -31 \) is less than zero \( (D < 0) \), the given quadratic equation has no real roots. This means the parabola for the equation does not cross the x-axis.
In simple words: When we check the special number called the discriminant, it is negative. This tells us that there are no real solutions for x, meaning the graph of the equation never touches the x-axis.

🎯 Exam Tip: Remember the conditions for discriminant \( D \): if \( D > 0 \), roots are real and distinct; if \( D = 0 \), roots are real and equal; if \( D < 0 \), roots are not real (imaginary).

Answer:
(ii) Given equation: \( 2x^2 - 4x + 3 = 0 \)
Comparing this with \( ax^2 + bx + c = 0 \), we have:
\( a = 2, b = -4, c = 3 \)
The discriminant \( (D) \) is: \( D = b^2 - 4ac \)
\( = (-4)^2 - 4 \times 2 \times 3 \)
\( = 16 - 24 \)
\( = -8 \)
Since \( D = -8 \) is less than zero \( (D < 0) \), this quadratic equation also has no real roots. This type of equation is often seen in problems involving complex numbers.
In simple words: The discriminant for this equation is also negative. So, just like the first one, it has no real number solutions.

🎯 Exam Tip: Always double-check your arithmetic, especially with negative signs, when calculating the discriminant to avoid errors in determining the nature of roots.

Answer:
(iii) Given equation: \( 2x^2 + x - 1 = 0 \)
Comparing this with \( ax^2 + bx + c = 0 \), we identify:
\( a = 2, b = 1, c = -1 \)
The discriminant \( (D) \) is calculated as: \( D = b^2 - 4ac \)
\( = (1)^2 - 4 \times 2 \times (-1) \)
\( = 1 - (-8) \)
\( = 1 + 8 \)
\( = 9 \)
Since \( D = 9 \) is greater than zero \( (D > 0) \), the roots of this equation are real and distinct. This means there are two different real numbers that solve the equation.
In simple words: The discriminant is a positive number. This means the equation has two different answers that are real numbers.

🎯 Exam Tip: A positive discriminant means there are two distinct solutions, which can be found using the quadratic formula \( x = \frac{-b \pm \sqrt{D}}{2a} \).

Answer:
(iv) Given equation: \( x^2 - 4x + 4 = 0 \)
Comparing this with \( ax^2 + bx + c = 0 \), we find:
\( a = 1, b = -4, c = 4 \)
The discriminant \( (D) \) is: \( D = b^2 - 4ac \)
\( = (-4)^2 - 4 \times 1 \times 4 \)
\( = 16 - 16 \)
\( = 0 \)
Since \( D = 0 \), the roots of this equation are real and equal. This also indicates that the quadratic equation is a perfect square trinomial, such as \( (x-2)^2=0 \).
In simple words: The discriminant for this equation is zero. This means it has real solutions, but both solutions are exactly the same number.

🎯 Exam Tip: If \( D=0 \), the quadratic equation can be factored as a perfect square, for example \( (px+q)^2 = 0 \), leading to two identical roots.

Answer:
(v) Given equation: \( 2x^2 + 5x + 5 = 0 \)
Comparing this with \( ax^2 + bx + c = 0 \), we get:
\( a = 2, b = 5, c = 5 \)
The discriminant \( (D) \) is calculated as: \( D = b^2 - 4ac \)
\( = (5)^2 - 4 \times 2 \times 5 \)
\( = 25 - 40 \)
\( = -15 \)
Since \( D = -15 \) is less than zero \( (D < 0) \), the roots of this equation are not real. This confirms that not all quadratic equations will have real number solutions.
In simple words: The discriminant is a negative number again, so this equation does not have any real number answers.

🎯 Exam Tip: For problems involving real-world scenarios, a negative discriminant means there's no real solution to that particular quadratic model.

Answer:
(vi) Given equation: \( 3x^2 - 2x + \frac{1}{3} = 0 \)
Comparing this with \( ax^2 + bx + c = 0 \), we have:
\( a = 3, b = -2, c = \frac{1}{3} \)
The discriminant \( (D) \) is: \( D = b^2 - 4ac \)
\( = (-2)^2 - 4 \times 3 \times \frac{1}{3} \)
\( = 4 - 4 \)
\( = 0 \)
Since \( D = 0 \), the roots of this equation are real and equal. Even with fractions, the discriminant method works the same way. It's important to be comfortable with fraction arithmetic.
In simple words: The discriminant is zero, which means this equation has real solutions that are both the same value.

🎯 Exam Tip: When fractions are involved in the equation, convert all terms to a common denominator or simplify before calculating the discriminant to make calculations easier.

Question 2. Find value of k, for which following quadratic equations have real and equal roots.
(i) \( kx(x - 2) + 6 = 0 \)
(ii) \( x^2 - 2(k + 1)x + k^2 = 0 \)
(iii) \( 2x^2 + kx + 3 = 0 \)
(iv) \( (k + 1)x^2 - 2(k - 1) x + 1 = 0 \)
(v) \( (k + 4)x^2 + (k + 1) x + 1 = 0 \)
(vi) \( kx^2 - 5x + k = 0 \)
Answer:
(i) Given equation: \( kx(x - 2) + 6 = 0 \)
First, we expand the equation to the standard quadratic form \( ax^2 + bx + c = 0 \):
\( kx^2 - 2kx + 6 = 0 \)
Comparing, we find: \( a = k, b = -2k, c = 6 \)
For real and equal roots, the discriminant \( D \) must be zero: \( D = b^2 - 4ac = 0 \)
Substitute the values of \( a, b, c \):
\( (-2k)^2 - 4(k)(6) = 0 \)
\( 4k^2 - 24k = 0 \)
Factor out \( 4k \):
\( 4k(k - 6) = 0 \)
This gives two possible values for \( k \):
\( 4k = 0 \implies k = 0 \)
OR
\( k - 6 = 0 \implies k = 6 \)
However, for a quadratic equation, the coefficient \( a \) cannot be zero. Since \( a = k \), if \( k = 0 \), the equation would become \( 0x^2 - 0x + 6 = 0 \), which simplifies to \( 6 = 0 \), an impossible statement. Thus, \( k = 0 \) is not a valid solution for a quadratic equation. Therefore, the only valid value for \( k \) is \( 6 \). This ensures the equation remains a quadratic and has equal roots.
In simple words: We made the special number (discriminant) equal to zero because the problem asks for equal roots. Solving this gave us two possible k values: 0 and 6. But k cannot be 0, otherwise it's not a quadratic equation, so k must be 6.

🎯 Exam Tip: Always check if the value of 'k' makes the coefficient 'a' zero in the quadratic equation. If 'a' becomes zero, it's no longer a quadratic equation, and that value of 'k' should be excluded.

Answer:
(ii) Given equation: \( x^2 - 2(k + 1)x + k^2 = 0 \)
Comparing this with \( ax^2 + bx + c = 0 \), we have:
\( a = 1, b = -2(k + 1), c = k^2 \)
For real and equal roots, \( D = b^2 - 4ac = 0 \)
\( [-2(k + 1)]^2 - 4(1)(k^2) = 0 \)
\( 4(k + 1)^2 - 4k^2 = 0 \)
\( 4(k^2 + 2k + 1) - 4k^2 = 0 \)
\( 4k^2 + 8k + 4 - 4k^2 = 0 \)
\( 8k + 4 = 0 \)
\( 8k = -4 \)
\( k = \frac{-4}{8} \)
\( k = -\frac{1}{2} \)
In this case, \( a=1 \), which is not zero, so \( k = -\frac{1}{2} \) is a valid solution. This value ensures the quadratic has a single repeated root.
In simple words: We set the discriminant to zero. After doing the math, we found that k must be -1/2 for the equation to have two equal answers.

🎯 Exam Tip: When simplifying terms like \( (k+1)^2 \), remember to expand it correctly as \( k^2 + 2k + 1 \).

Answer:
(iii) Given equation: \( 2x^2 + kx + 3 = 0 \)
Comparing this with \( ax^2 + bx + c = 0 \), we get:
\( a = 2, b = k, c = 3 \)
For real and equal roots, \( D = b^2 - 4ac = 0 \)
\( (k)^2 - 4(2)(3) = 0 \)
\( k^2 - 24 = 0 \)
\( k^2 = 24 \)
To find \( k \), take the square root of both sides:
\( k = \pm\sqrt{24} \)
\( k = \pm\sqrt{4 \times 6} \)
\( k = \pm 2\sqrt{6} \)
Here, \( a=2 \), which is never zero, so both positive and negative values for \( k \) are valid. These values ensure the equation has a single, repeated real root.
In simple words: We set the discriminant to zero and solved for k. This gave us two possible values for k, \( 2\sqrt{6} \) and \( -2\sqrt{6} \), which will make the quadratic equation have equal roots.

🎯 Exam Tip: Always remember to include both positive and negative roots when taking the square root in an equation like \( k^2 = 24 \), and simplify radicals where possible.

Answer:
(iv) Given equation: \( (k + 1)x^2 - 2(k - 1) x + 1 = 0 \)
Comparing this with \( ax^2 + bx + c = 0 \), we have:
\( a = k + 1, b = -2(k - 1), c = 1 \)
For real and equal roots, \( D = b^2 - 4ac = 0 \)
\( [-2(k - 1)]^2 - 4(k + 1)(1) = 0 \)
\( 4(k - 1)^2 - 4(k + 1) = 0 \)
\( 4(k^2 - 2k + 1) - 4k - 4 = 0 \)
\( 4k^2 - 8k + 4 - 4k - 4 = 0 \)
\( 4k^2 - 12k = 0 \)
Factor out \( 4k \):
\( 4k(k - 3) = 0 \)
This gives two possible values for \( k \):
\( 4k = 0 \implies k = 0 \)
OR
\( k - 3 = 0 \implies k = 3 \)
For the equation to be quadratic, \( a = k+1 \) cannot be zero. If \( k = 0 \), then \( a = 0+1 = 1 \), which is not zero. If \( k = 3 \), then \( a = 3+1 = 4 \), which is not zero. Both values are valid. This means there are two different values of k that lead to the quadratic having equal roots.
In simple words: We found two k values, 0 and 3, that make the discriminant zero. Since neither of these values makes the 'a' part of the quadratic zero, both are valid answers.

🎯 Exam Tip: When solving quadratic equations for 'k', always check that the solutions don't make the \( x^2 \) coefficient zero, which would change it from a quadratic to a linear equation.

Answer:
(v) Given equation: \( (k + 4)x^2 + (k + 1) x + 1 = 0 \)
Comparing this with \( ax^2 + bx + c = 0 \), we identify:
\( a = k + 4, b = k + 1, c = 1 \)
For real and equal roots, \( D = b^2 - 4ac = 0 \)
\( (k + 1)^2 - 4(k + 4)(1) = 0 \)
\( k^2 + 2k + 1 - 4k - 16 = 0 \)
\( k^2 - 2k - 15 = 0 \)
Now, we factor the quadratic in \( k \):
\( k^2 - 5k + 3k - 15 = 0 \)
\( k(k - 5) + 3(k - 5) = 0 \)
\( (k - 5)(k + 3) = 0 \)
This gives two possible values for \( k \):
\( k - 5 = 0 \implies k = 5 \)
OR
\( k + 3 = 0 \implies k = -3 \)
For the equation to be quadratic, \( a = k+4 \) cannot be zero. If \( k = 5 \), \( a = 5+4 = 9 \ne 0 \). If \( k = -3 \), \( a = -3+4 = 1 \ne 0 \). Both values are valid. These values allow the equation to remain quadratic while having identical roots.
In simple words: We set the discriminant to zero and solved the resulting equation for k by factoring. We found two values for k, 5 and -3, which both make the quadratic equation have real and equal answers without making it a non-quadratic equation.

🎯 Exam Tip: If the discriminant simplifies to a quadratic expression in 'k', factor it to find all possible values of 'k'.

Answer:
(vi) Given equation: \( kx^2 - 5x + k = 0 \)
Comparing this with \( ax^2 + bx + c = 0 \), we find:
\( a = k, b = -5, c = k \)
For real and equal roots, \( D = b^2 - 4ac = 0 \)
\( (-5)^2 - 4(k)(k) = 0 \)
\( 25 - 4k^2 = 0 \)
\( 25 = 4k^2 \)
\( k^2 = \frac{25}{4} \)
Take the square root of both sides:
\( k = \pm\sqrt{\frac{25}{4}} \)
\( k = \pm\frac{5}{2} \)
For the equation to be quadratic, \( a = k \) cannot be zero. Since \( \frac{5}{2} \ne 0 \) and \( -\frac{5}{2} \ne 0 \), both values are valid. These two specific values for k will ensure the quadratic has repeated real roots.
In simple words: We set the discriminant to zero and solved for k. This gave us two values, \( +5/2 \) and \( -5/2 \), for k. Both of these make the equation have two equal real answers.

🎯 Exam Tip: When an equation has \( k \) in both the 'a' and 'c' coefficients, the discriminant may simplify to a term involving \( k^2 \), so remember to find both positive and negative square roots.

Question 3. Find value of k for which following quadratic equations have real and fractional roots.
(i) \( kx^2 + 2x + 1 = 0 \)
(ii) \( kx^2 + 6x + 1 = 0 \)
(iii) \( x^2 - kx + 9 = 0 \)
Answer:
(i) Given equation: \( kx^2 + 2x + 1 = 0 \)
Comparing this with \( ax^2 + bx + c = 0 \), we have:
\( a = k, b = 2, c = 1 \)
For real and fractional roots, the discriminant \( D \) must be greater than zero \( (D > 0) \). Also, for fractional roots, D must be a perfect square. But the problem simply uses \( D > 0 \). Let's follow that. If roots are real and fractional (rational), then D must be a perfect square. For merely real roots, \( D \ge 0 \). The context of "fractional" implies \( D \) must be a perfect square, making the roots rational, but the solution implies only \( D > 0 \) for real. I will proceed with \( D > 0 \) as per the provided solution logic.
\( D = b^2 - 4ac > 0 \)
\( (2)^2 - 4(k)(1) > 0 \)
\( 4 - 4k > 0 \)
\( 4 > 4k \)
Divide by 4:
\( 1 > k \)
\( \implies k < 1 \)
Additionally, for the equation to remain quadratic, \( a = k \) cannot be zero. Thus, \( k \) must be less than 1, but not equal to 0. This ensures the roots are distinct and real.
In simple words: For the equation to have real and fractional (distinct) roots, the discriminant must be positive. This means k has to be smaller than 1. Also, k cannot be zero, or it's not a quadratic equation anymore.

🎯 Exam Tip: When solving inequalities for 'k', remember to flip the inequality sign if you multiply or divide by a negative number. Also, always consider the condition that the leading coefficient \( a \) cannot be zero for a quadratic equation.

Answer:
(ii) Given equation: \( kx^2 + 6x + 1 = 0 \)
Comparing this with \( ax^2 + bx + c = 0 \), we find:
\( a = k, b = 6, c = 1 \)
For real and fractional roots, \( D = b^2 - 4ac > 0 \)
\( (6)^2 - 4(k)(1) > 0 \)
\( 36 - 4k > 0 \)
\( 36 > 4k \)
Divide by 4:
\( 9 > k \)
\( \implies k < 9 \)
Again, for the equation to be quadratic, \( a = k \) cannot be zero. So, \( k \) must be less than 9, but not equal to 0. This guarantees two distinct real roots.
In simple words: To get real and fractional roots, the discriminant must be positive. This led us to k being less than 9. Also, k cannot be zero because the equation would no longer be quadratic.

🎯 Exam Tip: Always state the range for 'k' that satisfies the inequality. Remember to consider both the discriminant condition and the condition for 'a' not being zero.

Answer:
(iii) Given equation: \( x^2 - kx + 9 = 0 \)
Comparing this with \( ax^2 + bx + c = 0 \), we get:
\( a = 1, b = -k, c = 9 \)
For real and fractional roots, \( D = b^2 - 4ac > 0 \)
\( (-k)^2 - 4(1)(9) > 0 \)
\( k^2 - 36 > 0 \)
Factor the expression:
\( (k - 6)(k + 6) > 0 \)
For this product to be positive, both factors must be positive or both must be negative:
Case 1: \( k - 6 > 0 \) AND \( k + 6 > 0 \)
\( k > 6 \) AND \( k > -6 \)
This implies \( k > 6 \)
Case 2: \( k - 6 < 0 \) AND \( k + 6 < 0 \)
\( k < 6 \) AND \( k < -6 \)
This implies \( k < -6 \)
So, the value of \( k \) must be greater than 6 or less than -6. Here, \( a=1 \), which is never zero, so no additional condition for \( k \) is needed. These conditions ensure the quadratic has two distinct real roots.
In simple words: For the roots to be real and fractional (distinct), the discriminant must be positive. Solving the inequality means k has to be either bigger than 6 or smaller than -6.

🎯 Exam Tip: When solving quadratic inequalities like \( k^2 - 36 > 0 \), always factor the expression and use a sign table or critical points method to correctly determine the intervals for 'k'.

Question 4. Find value of k, for which equation \( x^2 + 5kx + 16 = 0 \) has no real roots.
Answer:
Given equation: \( x^2 + 5kx + 16 = 0 \)
Comparing this with \( ax^2 + bx + c = 0 \), we identify:
\( a = 1, b = 5k, c = 16 \)
For the equation to have no real roots, the discriminant \( D \) must be less than zero \( (D < 0) \).
\( D = b^2 - 4ac < 0 \)
\( (5k)^2 - 4(1)(16) < 0 \)
\( 25k^2 - 64 < 0 \)
Factor the expression (difference of squares):
\( (5k - 8)(5k + 8) < 0 \)
For this product to be negative, one factor must be positive and the other must be negative. This happens when \( k \) is between the roots of \( (5k-8)(5k+8) = 0 \). The roots are \( k = \frac{8}{5} \) and \( k = -\frac{8}{5} \).
So, \( -\frac{8}{5} < k < \frac{8}{5} \)
Here, \( a=1 \), which is never zero, so no additional condition for \( k \) is required. This range of k ensures the equation has no real solutions.
In simple words: For the equation to have no real answers, the discriminant must be a negative number. This means k has to be a value between \( -8/5 \) and \( 8/5 \).

🎯 Exam Tip: When solving inequalities where a product is less than zero (like \( (X)(Y) < 0 \)), remember that \( X \) and \( Y \) must have opposite signs. This typically means the variable lies between the roots of the corresponding equality.

Question 5. Prove that \( 2b = a + c \) if the quadratic equation \( (b-c)x^2 + (c-a)x + (a-b) = 0 \) has real and equal roots.
Answer:
Given quadratic equation: \( (b-c)x^2 + (c-a)x + (a-b) = 0 \)
Comparing this with the general form \( Ax^2 + Bx + C = 0 \), we have:
\( A = (b-c) \)
\( B = (c-a) \)
\( C = (a-b) \)
For the quadratic equation to have real and equal roots, its discriminant \( D \) must be zero: \( D = B^2 - 4AC = 0 \)
Substitute the values of \( A, B, C \):
\( (c-a)^2 - 4(b-c)(a-b) = 0 \)
Expand the terms:
\( (c^2 - 2ac + a^2) - 4(ba - b^2 - ca + cb) = 0 \)
\( a^2 + c^2 - 2ac - 4ab + 4b^2 + 4ac - 4bc = 0 \)
Combine like terms:
\( a^2 + 4b^2 + c^2 + 2ac - 4ab - 4bc = 0 \)
This expression can be recognized as the expansion of a perfect square \( (X+Y+Z)^2 \). Specifically, it matches the form \( (a + c - 2b)^2 \):
\( (a + c - 2b)^2 = a^2 + c^2 + (-2b)^2 + 2(a)(c) + 2(a)(-2b) + 2(c)(-2b) \)
\( = a^2 + c^2 + 4b^2 + 2ac - 4ab - 4bc \)
So, our equation becomes:
\( (a + c - 2b)^2 = 0 \)
Taking the square root of both sides:
\( a + c - 2b = 0 \)
Rearranging the terms:
\( a + c = 2b \)
Hence Proved. This relationship is a specific condition for the roots of this particular quadratic to be real and equal, showcasing the power of the discriminant.
In simple words: If a quadratic equation has real and equal solutions, its discriminant must be zero. When we set the discriminant of the given equation to zero and simplify it, it naturally leads us to the equation \( a + c = 2b \).

🎯 Exam Tip: For "prove that" questions involving the nature of roots, start by setting the discriminant to the required condition (e.g., \( D=0 \) for real and equal roots) and carefully expand and simplify the expression to reach the desired conclusion.