CAT — Quadratic Equations

Step 1: Rewrite the equation using the same base. Given: 9^(x² + 2x − 3) − 4(3^(x² + 2x − 2)) + 27 = 0. Since 9 = 3², 9^(x² + 2x − 3) = 3^[2(x² + 2x − 3)] = 3^(2x² + 4x − 6). Also, 3^(2x² + 4x − 6) = (3^(x² + 2x − 3))². Let y = 3^(x² + 2x − 3). Then, 3^(x² + 2x − 2) = 3 × 3^(x² + 2x − 3) = 3y. Substitute into the equation: y² − 4(3y) + 27 = 0 → y² − 12y + 27 = 0.

Step 2: Solve the quadratic equation. y² − 12y + 27 = 0 → (y − 3)(y − 9) = 0. Therefore, y = 3 or y = 9.

Step 3: Find the corresponding values of x. Since y = 3^(x² + 2x − 3). Case 1: y = 3. 3^(x² + 2x − 3) = 3 → x² + 2x − 3 = 1 → x² + 2x − 4 = 0. Using the quadratic formula: x = (−2 ± √20)/2 = −1 ± √5. Case 2: y = 9. 3^(x² + 2x − 3) = 9 = 3² → x² + 2x − 3 = 2 → x² + 2x − 5 = 0. Using the quadratic formula: x = (−2 ± √24)/2 = −1 ± √6.

Step 4: Find the product of all possible values of x. For x² + 2x − 4 = 0, product of roots = −4. For x² + 2x − 5 = 0, product of roots = −5. Hence, the product of all four values is (−4) × (−5) = 20.

Step 1: Find the minimum value of f(x). Since the coefficient of x² is positive, the parabola opens upward. The vertex is at x = 2c. Substituting: Minimum value of f(x) = (2c)² − 4c(2c) + 8c = 4c² − 8c² + 8c = −4c² + 8c.

Step 2: Find the maximum value of g(x). Since the coefficient of x² is negative, the parabola opens downward. The vertex is at x = 3c/2. Substituting: Maximum value of g(x) = −(3c/2)² + 3c(3c/2) − 2c = −9c²/4 + 9c²/2 − 2c = 9c²/4 − 2c.

Step 3: Apply the given condition. −4c² + 8c > 9c²/4 − 2c. Multiplying both sides by 4: −16c² + 32c > 9c² − 8c → 25c² − 40c < 0 → 5c(5c − 8) < 0. Therefore, 0 < c < 8/5. Among the options, only 1/2 lies in this range.

Given, log_x(x² + 12) = 4

Using the definition of logarithm, x⁴ = x² + 12

Rearrange, x⁴ − x² − 12 = 0

Let t = x²

Then, t² − t − 12 = 0

Factorizing, (t − 4)(t + 3) = 0

Since x is positive, t = 4

Therefore, x² = 4 → x = 2

Now, 3 log_y x = 1 → log_y x = 1/3

Using the definition of logarithm, y^(1/3) = x

Substitute x = 2, y^(1/3) = 2

Cubing both sides, y = 8

Therefore, x + y = 2 + 8 = 10

Answer: C. 10

Given, √(5x + 9) + √(5x − 9) = 3(2 + √2)
Square both sides.
LHS = (√(5x + 9) + √(5x − 9))² = (5x + 9) + (5x − 9) + 2√[(5x + 9)(5x − 9)] = 10x + 2√(25x² − 81)
RHS = [3(2 + √2)]² = 9(2 + √2)² = 9(4 + 4√2 + 2) = 54 + 36√2
Therefore, 10x + 2√(25x² − 81) = 54 + 36√2
Divide by 2, 5x + √(25x² − 81) = 27 + 18√2
Since 18√2 is the irrational part, √(25x² − 81) = 18√2 and 5x = 27
Therefore, x = 27/5
Now find, √(10x + 9) = √[10 × (27/5) + 9] = √(54 + 9) = √63 = 3√7
Answer:
C. 3√7

Step 1: Split the expression based on the absolute value. Given, x² − |x + 9| + x > 0. The absolute value changes at x = −9. So, consider two cases.

Case 1: x ≥ −9. Here, |x + 9| = x + 9. Substituting: x² − (x + 9) + x > 0 → x² − 9 > 0 → (x − 3)(x + 3) > 0. This is positive when x < −3 or x > 3. Since this case requires x ≥ −9, the solution becomes: −9 ≤ x < −3 or x > 3, i.e., [−9, −3) ∪ (3, ∞).

Case 2: x < −9. Here, |x + 9| = −(x + 9). Substituting: x² − [−(x + 9)] + x > 0 → x² + x + 9 + x > 0 → x² + 2x + 9 > 0. Completing the square: (x + 1)² + 8 > 0. This expression is always positive for every real value of x. Since this case requires x < −9, all values x < −9 satisfy the inequality. So the solution is: (−∞, −9).

Step 3: Combine both cases. (−∞, −9) ∪ [−9, −3) ∪ (3, ∞). Since x = −9 satisfies the inequality, (−∞, −9) ∪ [−9, −3) = (−∞, −3). Therefore, the complete solution set is (−∞, −3) ∪ (3, ∞).

Step 1: Let the common negative root be r.
r² + mr + 9 = 0
r² + nr + 17 = 0
r² + (m + n)r + 35 = 0

Step 2: Express m and n in terms of r
m = −(r² + 9)/r
n = −(r² + 17)/r

Step 3: Use the third equation
r² + [(−(r² + 9)/r) + (−(r² + 17)/r)]r + 35 = 0
r² − (r² + 9) − (r² + 17) + 35 = 0
9 − r² = 0 → r² = 9
Since the common root is negative, r = −3

Step 4: Find m and n
m = −(9 + 9)/(−3) = 6
n = −(9 + 17)/(−3) = 26/3

Step 5: Compute 2m + 3n
= 2(6) + 3(26/3) = 12 + 26 = 38

Given, x² + (x − 2y − 1)² = −4y(x + y)
Expand the square, (x − 2y − 1)² = x² + 4y² + 4y + 1 − 4xy − 2x
Substitute into the given equation, x² + x² + 4y² + 4y + 1 − 4xy − 2x = −4xy − 4y²
The terms −4xy cancel from both sides.
So, 2x² + 4y² − 2x + 4y + 1 = −4y²
2x² + 8y² − 2x + 4y + 1 = 0
Now complete the squares.
2(x² − x) + 8(y² + y/2) + 1 = 0
= 2[(x − 1/2)² − 1/4] + 8[(y + 1/4)² − 1/16] + 1 = 0
= 2(x − 1/2)² − 1/2 + 8(y + 1/4)² − 1/2 + 1 = 0
Therefore, 2(x − 1/2)² + 8(y + 1/4)² = 0
Since both squared terms are non-negative, each must be zero.
Hence, x = 1/2, y = −1/4
Now, x − 2y = 1/2 − 2(−1/4) = 1/2 + 1/2 = 1
Answer:
B. 1

Step 1: Rewrite the equation
Let A = 2^(2x²) and B = 2^(x+15).
Then: 2^(4x²) = A², 2^(2x²+x+16) = 2AB, 2^(2x+30) = B²
The equation becomes: A² − 2AB + B² = 0

Step 2: Factorize
(A − B)² = 0 → A = B

Step 3: Equate the exponents
2x² = x + 15 → 2x² − x − 15 = 0

Step 4: Solve the quadratic
2x² − 6x + 5x − 15 = 0
2x(x − 3) + 5(x − 3) = 0
(x − 3)(2x + 5) = 0
x = 3 or x = −5/2

Step 5: Find the required sum
Sum = 3 + (−5/2) = 1/2

Given:
The roots α and β of
3x² + λx − 1 = 0
satisfy
1/α² + 1/β² = 15.
Find the value of
(α³ + β³)².

Step 1: Find the sum and product of the roots
For the quadratic,
α + β = −λ/3
αβ = −1/3

Step 2: Use the given condition
1/α² + 1/β²
= (α² + β²)/(αβ)²
= 15
Since
(αβ)² = (−1/3)² = 1/9,
we get
α² + β² = 15 × 1/9
= 5/3

Step 3: Find α + β
Using
α² + β² = (α + β)² − 2αβ,
5/3 = (α + β)² − 2(−1/3)
5/3 = (α + β)² + 2/3
(α + β)² = 1
Thus,
α + β = ±1

Step 4: Find α³ + β³
Using
α³ + β³ = (α + β)³ − 3αβ(α + β),
α³ + β³
= (α + β)³ + (α + β)
= (α + β)[(α + β)² + 1]
Since
(α + β)² = 1,
α³ + β³
= 2(α + β)

Step 5: Find the required value
(α³ + β³)²
= [2(α + β)]²
= 4(α + β)²
= 4 × 1
= 4

Answer:
B. 4

Given, x³ + (2r + 1)x² + (4r − 1)x + 2 = 0
One of the roots is −2.
Substitute x = −2.
(−2)³ + (2r + 1)(−2)² + (4r − 1)(−2) + 2 = 0
−8 + 4(2r + 1) − 2(4r − 1) + 2 = 0
−8 + 8r + 4 − 8r + 2 + 2 = 0
0 = 0
Hence, −2 is a root for every value of r.
Now divide the cubic by (x + 2).
Using synthetic division with −2:
-2 | 1 2r+1 4r−1 2
-2 -4r+2 -2
1 2r−1 1 0
So the remaining quadratic is x² + (2r − 1)x + 1 = 0
For the other two roots to be real, the discriminant must be non-negative.
Discriminant = (2r − 1)² − 4 = 4r² − 4r + 1 − 4 = 4r² − 4r − 3
Therefore, 4r² − 4r − 3 ≥ 0
Now test the non-negative integers.
For r = 0: 4(0)² − 4(0) − 3 = −3 < 0 → Not possible.
For r = 1: 4(1)² − 4(1) − 3 = −3 < 0 → Not possible.
For r = 2: 4(2)² − 4(2) − 3 = 16 − 8 − 3 = 5 > 0 → Possible.
Hence, the minimum non-negative integer value of r is 2
Answer: 2

Step 1: Express all bases as powers of 2
8 = 2³ → 1/8 = 2^(−3)
32768 = 2¹⁵ → 1/32768 = 2^(−15)
Substituting: (2^(−3))^k × (2^(−15))^(1/3) = 2^(−3) × (2^(−15))^(1/k)

Step 2: Simplify the exponents
2^(−3k) × 2^(−5) = 2^(−3) × 2^(−15/k)
2^(−3k − 5) = 2^(−3 − 15/k)

Step 3: Equate the exponents
−3k − 5 = −3 − 15/k
3k + 5 = 3 + 15/k
3k + 2 = 15/k

Step 4: Form and solve the quadratic equation
3k² + 2k = 15 → 3k² + 2k − 15 = 0
Sum of roots = −b/a = −2/3

Step 1: Let the common root be α. Let the other root of 3x² − 5x + p = 0 be β, and the other root of 2x² − 2x + q = 0 be γ. We have to find β + γ.

Step 2: Use the sum of roots. For 3x² − 5x + p = 0, sum of roots = 5/3. So, α + β = 5/3 → β = 5/3 − α. For 2x² − 2x + q = 0, sum of roots = 1. So, α + γ = 1 → γ = 1 − α. Therefore, β + γ = (5/3 − α) + (1 − α) = 8/3 − 2α.

Step 3: Find the common root α. Since α satisfies both equations: 3α² − 5α + p = 0 and 2α² − 2α + q = 0. Multiply the second equation by 3: 6α² − 6α + 3q = 0. Multiply the first equation by 2: 6α² − 10α + 2p = 0. Subtract: −4α + 2p − 3q = 0 → 4α = 2p − 3q → α = p/2 − 3q/4.

Step 4: Substitute α into β + γ. β + γ = 8/3 − 2α = 8/3 − 2(p/2 − 3q/4) = 8/3 − p + 3q/2.

Step 1: Rewrite the quadratic Expand: x² − 2x + 1 + 2kx + 11 = 0 → x² + (2k − 2)x + 12 = 0 Step 2: Use the condition for no real roots Discriminant < 0: (2k − 2)² − 48 < 0 (k − 1)² < 12 −2√3 < k − 1 < 2√3 Since 2√3 ≈ 3.464: k < 4.464 Largest integer: k = 4 Step 3: Minimize the expression k/(4y) + 9y = 1/y + 9y By AM-GM: 1/y + 9y ≥ 2√(1/y × 9y) = 2√9 = 6 Equality when 1/y = 9y → y = 1/3 Least possible value = 6

Since the base 1/4 < 1, log₁/₄(x) > 0 implies 0 < x < 1. So we need 0 < n² − 7n + 11 < 1. From n² − 7n + 11 < 1: n² − 7n + 10 < 0 → (n−2)(n−5) < 0 → 2 < n < 5. Integer candidates: n = 3, 4. Checking domain (n² − 7n + 11 > 0): for n = 3: 9 − 21 + 11 = −1 < 0. For n = 4: 16 − 28 + 11 = −1 < 0. Neither satisfies the domain condition. Answer: 0.

Given, 2x² − 6x + k = 0
Let its roots be α and β.
Using Vieta's formulas, α + β = 6/2 = 3, αβ = k/2
Now, (α + β) and αβ are the roots of x² + px + p = 0
Therefore, Roots are 3 and k/2.
Again using Vieta's formulas,
Sum of roots = −p → 3 + k/2 = −p ...(1)
Product of roots = p → 3 × k/2 = p ...(2)
From (2), p = 3k/2
Substitute into (1), 3 + k/2 = −3k/2
Multiply throughout by 2, 6 + k = −3k → 4k = −6 → k = −3/2
Now, p = 3k/2 = 3(−3/2)/2 = −9/4
Finally, 8(k − p) = 8[−3/2 − (−9/4)] = 8(−6/4 + 9/4) = 8 × 3/4 = 6
Answer: 6

Step 1: Square both sides: f(x) + 1 = 9701²
f(x) = 9701² − 1 = 9700 × 9702

Step 2: Let y = x² + 3x.
y(y + 2) = 9700 × 9702
y² + 2y + 1 = 9701²
(y + 1)² = 9701²
y + 1 = ±9701 → y = 9700 or y = −9702

Step 3: Solve for x
Case 1: x² + 3x − 9700 = 0 → Discriminant = 9 + 38800 = 38809 = 197²
x = (−3 ± 197)/2 → x = 97 or x = −100

Case 2: x² + 3x + 9702 = 0 → Discriminant = 9 − 38808 < 0 (no real roots)

Step 4: Sum of real roots = 97 + (−100) = −3

Step 1: Let y = (2x − 3)/(2x² + 4x − 6).
Rearrange: 2yx² + (4y − 2)x + (3 − 6y) = 0

Step 2: For real x, discriminant ≥ 0.
(4y − 2)² − 4(2y)(3 − 6y) ≥ 0
16y² − 16y + 4 − 24y + 48y² ≥ 0
64y² − 40y + 4 ≥ 0
16y² − 10y + 1 ≥ 0

Step 3: Factor: (8y − 1)(2y − 1) ≥ 0
This holds when y ≤ 1/8 or y ≥ 1/2.

Step 4: Range = (−∞, 1/8] ∪ [1/2, ∞)

Given:
10ˣ + 4/10ˣ = 81/2
Find the sum of all distinct real values of x.

Step 1: Substitute a variable
Let
10ˣ = t
where
t > 0.
Then,
t + 4/t = 81/2

Step 2: Form a quadratic equation
Multiply both sides by 2t,
2t² + 8 = 81t
2t² − 81t + 8 = 0

Step 3: Solve the quadratic
Factorizing,
(2t − 1)(t − 8) = 0
Hence,
t = 1/2
or
t = 8
Since
t = 10ˣ,
the corresponding values of x are
x = log₁₀(1/2)
and
x = log₁₀8

Step 4: Find the sum of the values
Sum
= log₁₀(1/2) + log₁₀8
= log₁₀(8/2)
= log₁₀4
= log₁₀(2²)
= 2 log₁₀2

Answer:
A. 2 log₁₀2

Step 1: Find x + 1/x
x² + 1/x² = (x + 1/x)² − 2 = 25 → (x + 1/x)² = 27 → x + 1/x = 3√3

Step 2: x³ + 1/x³ = (x + 1/x)³ − 3(x + 1/x)
= (3√3)³ − 3(3√3) = 81√3 − 9√3 = 72√3

Step 3: x⁵ + 1/x⁵ = (x³ + 1/x³)(x² + 1/x²) − (x + 1/x)
= 72√3 × 25 − 3√3 = 1800√3 − 3√3 = 1797√3

Step 4: x⁷ + 1/x⁷ = (x⁵ + 1/x⁵)(x² + 1/x²) − (x³ + 1/x³)
= 1797√3 × 25 − 72√3 = 44925√3 − 72√3 = 44853√3

Given:
A 300-litre container is partially filled with water.
The remaining part is filled with milk.
Then, a quantity of the mixture equal to twice the initial amount of water is removed and replaced with water.
The final solution contains 72% milk.
Find the amount of water initially poured.

Step 1: Let the initial amount of water be x litres.
Then,
Initial milk = 300 − x litres.

Step 2: Amount of mixture removed
The amount removed is
2x litres.
Since the mixture is uniform,
Milk removed
= (300 − x)/300 × 2x

Step 3: Milk remaining
Milk remaining
= (300 − x) − (300 − x)/300 × 2x
After adding water, only the amount of water changes.
Since the final solution contains 72% milk,
Milk remaining
= 72% of 300
= 216 litres.
Hence,
(300 − x) − (300 − x)/300 × 2x = 216

Step 4: Solve the equation
Factor out (300 − x),
(300 − x)(1 − 2x/300) = 216
(300 − x)(300 − 2x) = 64800
(300 − x)(150 − x) = 32400
Expanding,
45000 − 450x + x² = 32400
x² − 450x + 12600 = 0
(x − 30)(x − 420) = 0
Since x cannot exceed 300,
x = 30

Final Answer
The amount of water initially poured into the container was
30 litres

Step 1: Domain conditions
x − 3 > 0 and x − 3 ≠ 1 → x > 3 and x ≠ 4.

Step 2: Rewrite using 2 = log₍ₓ₋₃₎(x−3)²
log₍ₓ₋₃₎(x² − 9) = log₍ₓ₋₃₎[(x+1)(x−3)²]
→ x² − 9 = (x+1)(x−3)²

Step 3: Factor and simplify
(x−3)(x+3) = (x+1)(x−3)²
Cancel (x−3) [valid since x > 3]: x + 3 = (x+1)(x−3)

Step 4: Expand and solve
x + 3 = x² − 2x − 3 → x² − 3x − 6 = 0
x = (3 ± √33)/2

Step 5: Check domain
(3 − √33)/2 < 3 → rejected.
Valid solution: x = (3 + √33)/2

Given:
For every non-zero real number x,
f(x) + 2f(1/x) = 3x
Find the sum of all possible values of x for which
f(x) = 3.

Step 1: Form another equation
Replace x by 1/x.
Then,
f(1/x) + 2f(x) = 3/x
We now have the two equations:
f(x) + 2f(1/x) = 3x
2f(x) + f(1/x) = 3/x

Step 2: Solve for f(x)
Multiply the first equation by 2,
2f(x) + 4f(1/x) = 6x
Subtract the second equation,
3f(1/x) = 6x − 3/x
f(1/x) = 2x − 1/x
Substitute into the first equation,
f(x) + 2(2x − 1/x) = 3x
f(x) = 3x − 4x + 2/x
f(x) = 2/x − x

Step 3: Use the given condition
Since
f(x) = 3,
2/x − x = 3
Multiply throughout by x,
2 − x² = 3x
x² + 3x − 2 = 0

Step 4: Find the sum of the solutions
For the quadratic,
x² + 3x − 2 = 0,
the sum of the roots is
−3

Final Answer
Answer:
C. −3

Step 1: Let the first term and common ratio be a and r. Then a₁ = a, a₂ = ar, a₃ = ar². Since the GP is decreasing and infinite, 0 < r < 1.

Step 2: Form the given equations. From the first condition: a + ar + ar² = 52 → a(1 + r + r²) = 52. Therefore, a = 52/(1 + r + r²). Also, a₁a₂ + a₂a₃ + a₃a₁ = 624 → a²r + a²r² + a²r³ = 624 → a²(r + r² + r³) = 624.

Step 3: Substitute the value of a. [52²/(1 + r + r²)²] × (r + r² + r³) = 624. Since r + r² + r³ = r(1 + r + r²): 52²r/(1 + r + r²) = 624. Since 52² = 2704: 2704r/(1 + r + r²) = 624. Dividing both sides by 208: 13r/(1 + r + r²) = 3.

Step 4: Solve for r. 13r = 3(1 + r + r²) → 13r = 3 + 3r + 3r² → 3r² − 10r + 3 = 0 → (3r − 1)(r − 3) = 0. So r = 1/3 or r = 3. Since the GP is decreasing, r = 1/3.

Step 5: Find the first term. a = 52/(1 + 1/3 + 1/9) = 52/(13/9) = 36.

Step 6: Find the sum of the infinite GP. Sum = a/(1 − r) = 36/(1 − 1/3) = 36/(2/3) = 54.

Given:
√(x + 6√2) − √(x − 6√2) = 2√2
Find x.

Step 1: Square both sides
[√(x + 6√2) − √(x − 6√2)]² = (2√2)²
x + 6√2 + x − 6√2 − 2√[(x + 6√2)(x − 6√2)] = 8
2x − 2√(x² − 72) = 8
Divide by 2,
x − √(x² − 72) = 4

Step 2: Isolate the square root
√(x² − 72) = x − 4
Since the left side is non-negative,
x ≥ 4.

Step 3: Square again
x² − 72 = (x − 4)²
x² − 72 = x² − 8x + 16
8x = 88
x = 11

Step 4: Verify the solution
Substituting x = 11,
√(11 + 6√2) − √(11 − 6√2)
= √9 − √1
= 3 − 1
= 2
= 2√2/√2
Since
11 + 6√2 = (3 + √2)²
and
11 − 6√2 = (3 − √2)²,
we have
√(11 + 6√2) = 3 + √2
√(11 − 6√2) = 3 − √2
Therefore,
(3 + √2) − (3 − √2)
= 2√2,
which satisfies the given equation.
Hence,
x = 11

Answer: 11

Given:
● Direct paths between P and Q = 3
● Direct paths between Q and S = 4
● Direct paths between P and R = 4
● No direct path between P and S
● Let the number of direct paths between Q and R = x
● Let the number of direct paths between R and S = y
Also,
● Number of ways to travel from P to S is 62.
● Number of ways to travel from Q to R is 27.
Find x.

Step 1: Form the equation for travel from P to S
Travel is possible in three ways.
Via Q
Ways = 3 × 4 = 12
Via R
Ways = 4 × y = 4y
Via Q and then R
Ways = 3 × x × y = 3xy
Hence,
12 + 4y + 3xy = 62
or,
4y + 3xy = 50
y(4 + 3x) = 50

Step 2: Form the equation for travel from Q to R
Travel is possible in three ways.
Directly
Ways = x
Via P
Ways = 3 × 4 = 12
Via S
Ways = 4 × y = 4y
Hence,
x + 12 + 4y = 27
or,
x + 4y = 15

Step 3: Solve the equations
From the second equation,
4y = 15 − x
Substitute into the first equation,
((15 − x)/4)(4 + 3x) = 50
Multiply by 4,
(15 − x)(4 + 3x) = 200
Expand,
60 + 45x − 4x − 3x² = 200
3x² − 41x + 140 = 0
Factorizing,
(3x − 20)(x − 7) = 0
Thus,
x = 20/3 or x = 7
Since the number of paths must be an integer,
x = 7

Final Answer
The number of direct paths between Q and R is
7

Let roots be r₁ and r₂. By Vieta's: r₁+r₂ = 5 and r₁r₂ = k ≥ 0. Integer pairs summing to 5: (0,5), (1,4), (2,3) and their reverses. Distinct products: 0×5 = 0, 1×4 = 4, 2×3 = 6. So k ∈ {0, 4, 6} → 3 values.

From 2pq − 20 = 52 − 2pq:
4pq = 72 → pq = 18
k = 2(18) − 20 = 16

From p² + q² − 29 = 16: p² + q² = 45

(p + q)² = p² + q² + 2pq = 45 + 36 = 81 → p + q = ±9
(p − q)² = p² + q² − 2pq = 45 − 36 = 9 → p − q = ±3

Possible pairs:
(p + q = 9, p − q = 3) → p = 6, q = 3 → p³ − q³ = 216 − 27 = 189
(p + q = 9, p − q = −3) → p = 3, q = 6 → p³ − q³ = 27 − 216 = −189
(p + q = −9, p − q = 3) → p = −3, q = −6 → p³ − q³ = −27 − (−216) = 189
(p + q = −9, p − q = −3) → p = −6, q = −3 → p³ − q³ = −216 − (−27) = −189

Maximum = 189, Minimum = −189
Difference = 189 − (−189) = 378

Step 1: Square both sides
(a + b√n)² = 29 − 12√5
a² + b²n + 2ab√n = 29 − 12√5

Step 2: Compare rational and irrational parts
a² + b²n = 29
ab√n = −6√5

Step 3: Eliminate √n
Squaring: a²b²n = 180
Using b²n = 29 − a²:
a²(29 − a²) = 180 → a⁴ − 29a² + 180 = 0
(a² − 20)(a² − 9) = 0
Since a is an integer: a² = 9 → a = ±3

Step 4: Find b²n
b²n = 29 − 9 = 20
Possible values: b² = 1, n = 20 or b² = 4, n = 5

Step 5: Check each case
Case 1: b² = 1, n = 20 → ab = −3
a = −3, b = 1: −3 + √20 > 0 ✓ → a + b + n = −3 + 1 + 20 = 18

Case 2: b² = 4, n = 5 → ab = −6
a = −3, b = 2: −3 + 2√5 > 0 ✓ → a + b + n = −3 + 2 + 5 = 4

Step 6: Maximum value = 18

Given:
After two successive salary increments, Gopal's salary becomes
187.5% of the initial salary.
The second percentage increase is twice the first.
Find the percentage increase in the first increment.

Step 1: Let the first increment be x%.
Then the second increment is
2x%.
The final salary is
187.5% = 15/8
of the initial salary.
Hence,
(1 + x/100)(1 + 2x/100) = 15/8

Step 2: Form the quadratic equation
Multiplying both sides by 10000,
(100 + x)(100 + 2x) = 18750
Expanding,
10000 + 300x + 2x² = 18750
2x² + 300x − 8750 = 0
Divide throughout by 2,
x² + 150x − 4375 = 0

Step 3: Solve the quadratic
Factorizing,
(x + 175)(x − 25) = 0
So,
x = 25
or
x = −175
Since the increment cannot be negative,
x = 25

Final Answer
The percentage increase in the first increment is
25%

Answer:
C. 25