Concept of Quadratic Equations | Hard
The equations 3x² − 5x + p = 0 and 2x² − 2x + q = 0 have one common root.
The sum of the other roots of these two equations is
A) 1/3 − 2p + 3/2 q
B) 2/3 + p + 3/4 q
C) 5/3 − p + 3/2 q
D) 4/3 − p + 2/3 q
Answer & Explanation
Correct Answer: Option C
Given equations:
3x² − 5x + p = 0
2x² − 2x + q = 0
For a quadratic ax² + bx + c = 0, sum of roots = −b/a.
Step 1: Use the fact that r is a common root
Since r satisfies both equations:
From (1): 3r² − 5r + p = 0 → p = 5r − 3r²
From (2): 2r² − 2r + q = 0 → q = 2r − 2r²
Step 2: Express sum of the “other” roots
Equation (1): Sum of roots = 5/3. So, other root of (1) = 5/3 − r
Equation (2): Sum of roots = 1. So, other root of (2) = 1 − r
Required sum = (5/3 − r) + (1 − r) = 8/3 − 2r
Step 3: Eliminate r using p and q
From p = 5r − 3r²; From q = 2r − 2r²
Rewrite both: p = r(5 − 3r); q = r(2 − 2r)
Solve for r in linear combination form:
Multiply second by 3/2: (3/2)q = 3r − 3r²
Now subtract from p: p − (3/2)q = (5r − 3r²) − (3r − 3r²) = 2r
So: r = (p − 3q/2)/2 = p/2 − 3q/4
Step 4: Substitute r into required sum Required sum = 8/3 − 2r = 8/3 − 2(p/2 − 3q/4) = 8/3 − p + 3q/2.
Concepts of Polynomial Equations + Quadratic | Hard
If 9(x² + 2x – 3) − 4(3(x² + 2x − 2)) + 27 = 0,
then the product of all possible values of x is _____
A) 20
B) 15
C) 30
D) 5
Answer & Explanation
Correct Answer: A) 20
Given equation:
9^(x² + 2x − 3) − 4·3^(x² + 2x − 2) + 27 = 0
Step 1: Write everything with base 3
9 = 3², so ⇒ 9^(x² + 2x − 3) = 3^(2x² + 4x − 6)
Also, 3^(x² + 2x − 2) = 3·3^(x² + 2x − 3)
Let t = 3^(x² + 2x − 3), where t > 0.
Then the equation becomes: t² − 4(3t) + 27 = 0 ⇒ t² − 12t + 27 = 0
Step 2: Solve the quadratic in t
t² − 12t + 27 = 0 ⇒ (t − 3)(t − 9) = 0 ⇒ So t = 3 or t = 9
Step 3: Convert back to x
Case 1: 3^(x² + 2x − 3) = 3 ⇒ x² + 2x − 3 = 1 ⇒ x² + 2x − 4 = 0
Roots: x = −1 ± √5 ⇒ Product of roots = −4
Case 2: 3^(x² + 2x − 3) = 9 = 3² ⇒ x² + 2x − 3 = 2 ⇒ x² + 2x − 5 = 0
Roots: x = −1 ± √6 ⇒ Product of roots = −5
Step 4: Product of all possible values of x
Total product = (product from case 1) × (product from case 2)
= (−4) × (−5) = 20
Vedic Patterns G Strategy | Quadratic Equations | Hard
If m and n are integers such that (m + 2n)(2m + n) = 27,
then the maximum possible value of 2m − 3n is ___
Answer & Explanation
Given: (m + 2n)(2m + n) = 27
Since m and n are integers, (m + 2n) and (2m + n) must be integer factors of 27.
Possible factor pairs (m + 2n, 2m + n) are:
(1, 27), (3, 9), (9, 3), (27, 1),
(−1, −27), (−3, −9), (−9, −3), (−27, −1)
We now solve each pair as a system: m + 2n = A; 2m + n = B
From these:
Multiply first equation by 2: 2m + 4n = 2A
Subtract from second: (2m + n) − (2m + 4n) = B − 2A
⇒ −3n = B − 2A ⇒ n = (2A − B) / 3
Then m = A − 2n
| m + 2n | 2m + n | n = (2A − B)/3 | m = A − 2n | 2m − 3n | Valid / Reject |
| 1 | 27 | (2 − 27)/3 = −25/3 | — | — | Reject |
| 3 | 9 | (6 − 9)/3 = −1 | 3 − 2(−1) = 5 | 13 | Valid |
| 9 | 3 | (18 − 3)/3 = 5 | 9 − 10 = −1 | −17 | Valid |
| 27 | 1 | (54 − 1)/3 = 53/3 | — | — | Reject |
| −1 | −27 | (−2 + 27)/3 = 25/3 | — | — | Reject |
| −3 | −9 | (−6 + 9)/3 = 1 | −3 − 2 = −5 | −13 | Valid |
| −9 | −3 | (−18 + 3)/3 = −5 | −9 + 10 = 1 | 17 | Valid |
| −27 | −1 | (−54 + 1)/3 = −53/3 | — | — | Reject |
Maximum possible valid value of 2m − 3n = 17
X maro G Strategy | Inequality | Easy
The set of all real values of x for which (x² − |x + 9| + x) > 0, is
A) (−∞, −3) ∪ (3, ∞)
B) (−9, −3) ∪ (3, ∞)
C) (−∞, −9) ∪ (3, ∞)
D) (−∞, −9) ∪ (9, ∞)
Answer & Explanation
Use X maro G Strategy put x = -5 and x = -10
| Option | Interval | Test x = −5 | Test x = −10 | Status |
| A | (−∞, −3) ∪ (3, ∞) | Included, LHS = 16 > 0 | Included, LHS = 89 > 0 | ✅ Possible |
| B | (−9, −3) ∪ (3, ∞) | Included, LHS = 16 > 0 | Not included | ❌ Eliminated |
| C | (−∞, −9) ∪ (3, ∞) | Not included | Included, LHS = 89 > 0 | ❌ Eliminated |
| D | (−∞, −9) ∪ (9, ∞) | Not included | Included, LHS = 89 > 0 | ❌ Eliminated |
Maxima Minima G Strategy | Easy
If a, b, c and d are integers such that their sum is 46, then the minimum possible value of
(a − b)² + (a − c)² + (a − d)² is ______
Answer & Explanation
Minimum possible value is 2
Since a, b, c, d must be integers, they must be the integers closest to 11.5, which are 11 and 12.
To satisfy the sum constraint a + b + c + d = 46, we can take values 11, 11, 12, 12.
Expression becomes: (11 − 12)² + (11 − 12)² + (11 − 11)² = 1 + 1 + 0 = 2
