Test: Quadratic Equations (Medium) - Class 10 MCQ

Another number = 27 – x

Product of two numbers = 182 x(27 – x) = 182

⇒ x² – 27x – 182 = 0

⇒ x² – 13x – 14x + 182 = 0

⇒ x(x – 13) -14(x – 13) = 0

⇒ (x – 13)(x -14) = 0

⇒ x = 13 or x = 14

(1 / 2)² – m(1 / 2) – 5 / 4 = 0

1 / 4 - m / 2 - 5 / 4 = 0

m = -2

Altitude = (x – 7) cm

In a right triangle,

Base² + Altitude² = Hypotenuse² (From Pythagoras theorem)

∴ x² + (x – 7)² = 132

By solving the above equation, we get;

⇒ x = 12 or x = – 5

Since the side of the triangle cannot be negative.

Therefore, base = 12cm and altitude = 12 - 7 = 5cm

⇒ 2x² + x = -4

Dividing the equation by 2, we get

⇒ x² + 1 / 2x = -2

⇒ x² + 2 × x × 1 / 4 = -2

By adding (1 / 4)² to both sides of the equation, we get

⇒ (x)² + 2 × x × 1 / 4 + (1 / 4)² = (1 / 4)² – 2

⇒ (x + 1 / 4)² = 1 / 16 – 2

⇒ (x + 1 / 4)² = -31 / 16

The square root of negative number is imaginary, therefore, there is no real root for the given equation.

Hence, we can write, √(6 + x) = x

6 + x = x²

x² - x - 6 = 0

x² - 3x + 2x - 6 = 0

x(x - 3) + 2(x - 3) = 0

(x + 2) (x - 3) = 0

x = -2, 3

Since, x cannot be negative, therefore, x = 3

Three years ago his age = x – 3

Five years later his age = x + 5

Given, the sum of the reciprocals of Rehman’s ages 3 years ago and after 5 years is equal to 1 / 3.

∴ 1 / x - 3 + 1 / x - 5 = 1 / 3

(x + 5 + x - 3) / (x - 3)(x + 5) = 1 / 3

(2x + 2) / (x - 3)(x + 5) = 1 / 3

⇒ 3(2x + 2) = (x - 3)(x + 5)

⇒ 6x + 6 = x² + 2x – 15

⇒ x² – 4x – 21 = 0

⇒ x² – 7x + 3x – 21 = 0

⇒ x(x – 7) + 3(x – 7) = 0

⇒ (x – 7)(x + 3) = 0

⇒ x = 7, -3

We know age cannot be negative, hence the answer is 7.

Time required to cover 360 km = 360/x hr.

As per the question given,

⇒ (x + 5)(360 - 1 / x) = 360

⇒ 360 – x + 1800 - 5 / x = 360

⇒ x² + 5x + 10x – 1800 = 0

⇒ x(x + 45) - 40(x + 45) = 0

⇒ (x + 45)(x – 40) = 0

⇒ x = 40, -45

Negative value is not considered for speed hence the answer is 40km/hr.

∴ D = 0 => b² – 4ac = 0

⇒ (c – a)² -4(b – c) (a – b) = 0

⇒ c² – b² – 2ac - 4(ab - b² + bc) = 0 => c + a - 2b = 0 => c + a = 2b

⇒ c² + a² – 2ca – 4ab + 4b^² + 4ac – 4bc = 0

⇒ c² + a² + 4b² + 2ca – 4ab – 4bc = 0

⇒ (c + a – 2b)² = 0 ⇒ c + a – 2b = 0

⇒ c + a = 2b

2x² + px -15 = 0

∴ 2(-5)² + p (-5) – 15 = 0

⇒ 50 – 5p -15 = 0

⇒ 5p = 35 ⇒ p = 7

Since the given equation has real and equal roots

∴ b² – 4ac = 0

⇒ (4k)² – 4 × 12 × 3 = 0

⇒ 16k² – 144 = 0

⇒ k² = 9

⇒ k = ±3

∴ D = b² – 4ac = (1)² – 4 × 2 × (-1) = 1 + 8 = 9 > 0

∴ Roots of the given equation are real and distinct.

∴ 2(2)² + k(2) – 6 = 0

⇒ 8 + 2k – 6 = 0

⇒ 2k = -2

∴ k = -1

∴ αβ = 1

⇒ 2 / p = 1

∴ p = 2

Correct product = -9 x -1 = 9 from Sohan

∴ x² – (10)x + 9 = 0

⇒ x² – 10x + 9 = 0

⇒ x² – 9x – x + 9

⇒ x(x – 9) – 1(x – 9) = 0

⇒ (x - 9) (x -1) = 0 .

⇒ Correct roots are 9 and 1.

⇒ (x – α) (x – β) – c = (x – a) (x – b)

This shows that roots of (x – α) (x – β) – c are a and b.

∴ Sum of the roots = −ba = −(−9)³ = 3

∴ b² – 4AC = 0

⇒ (k)² – 4 × 2 × 3 = 0

⇒ k² = 24

⇒ k = ± √24

∴ other root is 3 – √2

∴ Sum of roots = 3 + √2 + 3 – √2 = 6

Product of roots = (3 + √2)(3 – √2) = (3)² – (√2)² = 9 – 2 = 7

∴ Required quadratic equation is x² – 6x + 7 = 0

⇒ x² + 4 – 2 × x × 2 + 1 = 2x – 3

⇒ x² – 4x + 5 – 2x + 3 = 0

∴ x² – 6x + 8 = 0, which is a quadratic equation.

⇒ x² + x + 8 = x² – 4

⇒ x² + x + 8 - x² + 4 = 0

⇒ x + 12 = 0, which is a linear equation.