Math Olympiad Test: Quadratic Equations- 4 - Free MCQ with solutions

Since roots of the given equation are equal
∴ D = 0
⇒ (–2b(a + c))² – 4(a² + b²)(b² + c²) = 0
⇒ 4b²(a² + c² + 2ac) – 4(a²b² + a2c² + b⁴ + b²c²) = 0
⇒ a²b² + b²c² + 2ab²c – a²b² – a²c² – b⁴ – b²c² = 0
⇒ 2ab²c – a²c² – b⁴ = 0 ⇒ b⁴ + a²c² – 2ab²c = 0
⇒ (b² – ac)² = 0 ⇒ b² = ac

We have, 3x^1/2 + 5x^−1/2 = √2

⇒ 3x + 5 = (2x)^1/2 ...(1)
Squaring both sides of (1), we have
9x² + 25 + 30x = 2x
⇒ 9x² + 28x + 25 = 0

Given equation is x^2/3 + x^1/3 – 2 = 0
Let y = x^1/3 ⇒ y² + y – 2 = 0
⇒ y² + 2y – y – 2 = 0 ⇒ (y – 1)(y + 2) = 0
⇒ y = 1 or y = –2 ⇒ x^1/3 = 1 or x^1/3 = –2
⇒ x = 1 or x = – 8

Let the two roots be a and b, then a + b = 12 ...(1) and a – b = 4 ...(2)
⇒ a = 8 and b = 4 (from (1) and (2))
∴ Required equation is x² – 12x + 32 = 0

Let the number of bangles in a side of square = x
According to the question, x² + 38 = Total no. of bangles ...(1)
Also, (x + 1)² – 25 = Total no. of bangles ...(2)
From (1) and (2), we have
x² + 38 = (x + 1)² – 25
⇒ 38 + 24 = 2x ⇒ x = 31
∴ Total no. of bangles = (31)² + 38 = 999

We have, 
⇒ (x² – x)m – (m² + m) = x(x – 1)(m – 1)
⇒ x²m – mx – m² – m = x²m – x² – xm + x
⇒ (x² – m²) – (m + x) = 0
⇒ (x + m)(x – m – 1) = 0
Now, since roots are equal ⇒ –m = m + 1 ⇒ m = -(1/2)

For 1^st one,
Let the equation be x² + ax + b = 0
Since roots are 5 and 9
∴ a = –14 and b = 45
For 2^nd one,
Let the equation be x² + px + q = 0
Since roots are 12 and 4.
∴ p = –16 and q = 48
Now, according to the question, b and p both are wrong. Therefore, the correct equation would be x² –14x + 48 = 0

Let the speed of man be x km/hr and the time taken by him to cover 48 km with speed x be t
According to the question, 48 = x × t ...(1)
Also, 48 = (x + 2)(t – 4) ...(2)
⇒ 48 = xt – 4x + 2t – 8
⇒ 56 = 48 – 4x + 2t [Using (1)]
⇒ 8 = – 4x + 2t ⇒ 4 = –2x + t
⇒ 4 = −2x + (48/x) [Using (1)]
⇒ 2x = –x² + 24 ⇒ x² + 2x – 24 = 0
⇒ 
Since, speed cannot be negative
∴ Required speed is 4 km/hr.

Let the speed of the stream be x km/h
Speed of the boat in upstream = (5 - x)km/h
Speed of the boat in downstream = (5 + x)km/h
Time, say t₁ (in hours), for going 5.25 km upstream = 5.25/5 - x
Time, say t₂ (in hours), for returning 5.25 km downstream = 5.25/5 + x
Obviously t₁ > t₂
Therefore, according to the given condition of the problem,
t₁ = t₂ + 1

This gives x = 2, since we reject x = -25/2
Thus, the speed of the stream is 2 km/h.

Let x be the number of persons and y be the amount received by each person.
According to the question, xy = 6500 ...(1)
Also, (x + 15)(y – 30) = 6500 ...(2)
⇒ xy – 30x + 15y – 450 = 6500
⇒ 6500 – 30x + 15y – 450 = 6500 [From (1)]
⇒ 2x – y + 30 = 0
⇒ 2x - (6500/x) + 30 = 0 [From (1)]
⇒ 2x² – 6500 + 30x = 0
⇒ x² + 15x – 3250 = 0
⇒ (x – 50)(x + 65) = 0
⇒ x = 50 or x = – 65
Since, number of persons cannot be negative. Hence, number of persons = 50