Quadratic Equation with Conjugate Roots
Assertion (A): If 5 + √7 is a root of a quadratic equation with rational coefficients, then its other root is 5 – √7.
Reason (R): Surd roots of a quadratic equation with rational coefficients occur in conjugate pairs.
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✅ Answer & Explanation
Correct Option: Both assertion (A) and reason (R) are true, and reason (R) is the correct explanation of assertion (A).
Explanation: When a quadratic equation has rational coefficients, irrational roots always occur in conjugate pairs. If one root is 5 + √7 (an irrational number involving surd), the other root must be its conjugate 5 – √7 to ensure the coefficients remain rational when we form the equation from its roots.
The assertion states that if \(5+\sqrt{7}\) is a root of a quadratic equation with rational coefficients, then its other root is \(5-\sqrt{7}\). This is a direct application of the property that surd roots of quadratic equations with rational coefficients always appear as conjugate pairs (e.g., \(a+\sqrt{b}\) and \(a-\sqrt{b}\)). The reason correctly explains this property, making it the correct explanation for the assertion.
Zeroes Double in Value
If the zeroes of the polynomial x² + px + q are double in value to the zeroes of 2x² – 5x – 3, find the value of p and q.
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✅ Solution
Step 1: Find zeroes of 2x² – 5x – 3
Let zeroes be α and β
α + β = -(-5)/2 = 5/2
αβ = (-3)/2 = -3/2
Step 2: Zeroes of x² + px + q are 2α and 2β
Sum = 2α + 2β = 2(α+β) = 2 × (5/2) = 5
∴ -p = 5 ⇒ p = -5
Product = (2α)(2β) = 4αβ = 4 × (-3/2) = -6
∴ q = -6
Answer: p = -5, q = -6
Polynomial with Condition
If α, β are the zeroes of the polynomial f(x) = x² – p(x + 1) – c, such that (α + 1)(β + 1) = 0, then find c.
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✅ Solution
Step 1: Rewrite the polynomial
f(x) = x² – p(x + 1) – c
= x² – px – p – c
= x² – px – (p + c)
Step 2: Use sum and product of zeroes
α + β = p
αβ = -(p + c)
Step 3: Use given condition (α + 1)(β + 1) = 0
(α + 1)(β + 1) = αβ + α + β + 1 = 0
-(p + c) + p + 1 = 0
-p – c + p + 1 = 0
-c + 1 = 0
c = 1
Answer: c = 1
Find Quadratic Polynomial
Find a quadratic polynomial where zeroes are 5 – 3√2 and 5 + 3√2.
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✅ Solution
Step 1: Find sum of zeroes
Sum = (5 – 3√2) + (5 + 3√2) = 10
Step 2: Find product of zeroes
Product = (5 – 3√2)(5 + 3√2)
= 5² – (3√2)²
= 25 – 18
= 7
Step 3: Form quadratic polynomial
Quadratic polynomial: x² – (sum)x + product
= x² – 10x + 7
Answer: x² – 10x + 7 (or k(x² – 10x + 7) where k is any constant)
Find Value of k
If α and β are the zeroes of the polynomial f(x) = x² – 6x + k, find the value of k such that α² + β² = 40.
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✅ Solution
Step 1: Find sum and product of zeroes
For f(x) = x² – 6x + k:
α + β = 6
αβ = k
Step 2: Use identity for α² + β²
α² + β² = (α + β)² – 2αβ
40 = (6)² – 2k
40 = 36 – 2k
Step 3: Solve for k
2k = 36 – 40
2k = -4
k = -2
Answer: k = -2
Polynomial with Transformed Zeroes
If α and β are the zeroes of the quadratic polynomial f(x) = x² – x – 2, find a polynomial whose zeroes are 2α + 1 and 2β + 1.
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✅ Solution
Step 1: Find α and β from f(x) = x² – x – 2
α + β = 1
αβ = -2
Step 2: Find sum of new zeroes
Sum = (2α + 1) + (2β + 1)
= 2α + 2β + 2
= 2(α + β) + 2
= 2(1) + 2 = 4
Step 3: Find product of new zeroes
Product = (2α + 1)(2β + 1)
= 4αβ + 2α + 2β + 1
= 4αβ + 2(α + β) + 1
= 4(-2) + 2(1) + 1
= -8 + 2 + 1 = -5
Step 4: Form new polynomial
New polynomial: x² – (sum)x + product
= x² – 4x – 5
Answer: x² – 4x – 5
System of Equations (CBSE 2001)
Find the values of p and q for which the following system of equations has infinite number of solutions:
2x + 3y = 7
(p + q)x + (2p – q)y = 21
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✅ Solution
Step 1: Condition for infinite solutions
For infinite solutions: a₁/a₂ = b₁/b₂ = c₁/c₂
Here: a₁ = 2, b₁ = 3, c₁ = -7
a₂ = p+q, b₂ = 2p-q, c₂ = -21
Step 2: Apply conditions
1. 2/(p+q) = 3/(2p-q)
2(2p-q) = 3(p+q)
4p – 2q = 3p + 3q
p = 5q … (i)
2. 2/(p+q) = 7/21 = 1/3
2/(p+q) = 1/3
p+q = 6 … (ii)
Step 3: Solve equations
From (i) and (ii):
p = 5q and p + q = 6
5q + q = 6
6q = 6 ⇒ q = 1
p = 5(1) = 5
Answer: p = 5, q = 1
Age Problem (CBSE 2003)
Father’s age is three times the sum of ages of his two children. After 5 years his age will be twice the sum of ages of two children. Find the age of father.
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✅ Solution
Step 1: Let present ages
Let sum of children’s ages = x years
Father’s present age = 3x years
Step 2: After 5 years
After 5 years:
Father’s age = 3x + 5
Sum of children’s ages = x + 10 (each child ages 5 years, so total +10)
Step 3: Form equation
According to problem:
3x + 5 = 2(x + 10)
3x + 5 = 2x + 20
3x – 2x = 20 – 5
x = 15
Father’s age = 3x = 3 × 15 = 45 years
Answer: Father’s age = 45 years
Test Marks Problem (NCERT)
Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?
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✅ Solution
Step 1: Define variables
Let: Number of correct answers = x
Number of wrong answers = y
Total questions = x + y
Step 2: Form equations
Case 1: 3x – y = 40 … (i)
Case 2: 4x – 2y = 50 … (ii)
Step 3: Solve equations
Multiply (i) by 2: 6x – 2y = 80 … (iii)
Subtract (ii) from (iii):
(6x – 2y) – (4x – 2y) = 80 – 50
2x = 30 ⇒ x = 15
From (i): 3(15) – y = 40
45 – y = 40 ⇒ y = 5
Total questions = x + y = 15 + 5 = 20
Answer: 20 questions
Library Charges Problem
A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Shristi paid ₹27 for a book kept for seven days, while Rekha paid ₹21 for the book she kept for five days. Find the fixed charge and the additional charge.
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✅ Solution
Step 1: Define variables
Let: Fixed charge for 3 days = ₹x
Additional charge per day = ₹y
Step 2: Form equations
Shristi (7 days): Fixed 3 days + 4 extra days
x + 4y = 27 … (i)
Rekha (5 days): Fixed 3 days + 2 extra days
x + 2y = 21 … (ii)
Step 3: Solve equations
Subtract (ii) from (i):
(x + 4y) – (x + 2y) = 27 – 21
2y = 6 ⇒ y = 3
From (ii): x + 2(3) = 21
x + 6 = 21 ⇒ x = 15
Answer: Fixed charge = ₹15, Additional charge = ₹3 per day