ব্যাংক ডেইলি কুইজ [লং কোর্সের অংশ]

Question: Find the difference between the roots of the quadratic equation x² - 9x + 20 = 0.

Solution:
x² - 9x + 20 = 0
⇒ x² - 4x - 5x + 20 = 0
⇒ x(x - 4) - 5(x - 4) = 0
⇒ (x - 5)(x - 4) = 0

The solutions to the equation are 5 and 4.
Their difference is 1.

Question: Solve: 2y² = 13y + 45

Solution:
2y² = 13y + 45
⇒ 2y² - 13y - 45 = 0
⇒ 2y² - 18y + 5y - 45 = 0
⇒ 2y(y - 9) + 5(y - 9) = 0
⇒ (y - 9)(2y + 5) =0
∴ y - 9 = 0  or, 2y + 5 = 0
∴ y = 9        ∴ y = - 5/2

Question:

Solution:
(2x - 1)/(x - 1) = (7x - 1)/(2x + 2)
⇒ (2x - 1)(2x + 2) = (7x - 1)(x - 1)
⇒ 4x² + 4x - 2x - 2 = 7x² - 7x - x + 1
⇒ 4x² + 2x - 2 = 7x²- 8x + 1
⇒ 3x² - 10x + 3 = 0
⇒ 3x² - 9x - x + 3 = 0
⇒ 3x(x - 3) - 1(x - 3) = 0
⇒ (x - 3)(3x - 1) = 0
∴ x = 3 or x = 1/3

Question: Solve: 16y² = 32y³ + 2y

Solution:
16y² = 32y³ + 2y
⇒ 16y = 32y² + 2
⇒ 32y² - 16y + 2 = 0
⇒ 16y² - 8y + 1 = 0
⇒ (4y)² - 2.4y.1 + 1² = 0
⇒ (4y - 1)² = 0
⇒ 4y - 1 = 0
∴ y = 1/4

Question: If α, β are the roots of the equation x² - 7x + 12 = 0, then α² + β² equals to:

Solution:
x² - 7x + 12 = 0
⇒ x² - 3x - 4x + 12 = 0
⇒ x(x - 3) - 4(x - 3) = 0
⇒ (x - 3)(x - 4) = 0
∴ x = 3, 4

Hence, α = 3, β = 4
Hence, The value of α² + β² = 3² + 4² = 9 + 16 = 25

Question: For the function f(x) = x² + 2x - 2, find x when f(x) = 6.

Solution:
f(x) = x² + 2x - 2
f(x) = 6

∴ x² + 2x - 2 = 6
⇒ x² + 2x - 2 - 6 = 0
⇒ x² + 2x - 8 = 0
⇒ x² + 4x - 2x - 8 = 0
⇒ x(x + 4) - 2(x + 4) = 0
⇒ (x + 4)(x - 2) = 0
∴ x + 4 = 0   or  x - 2 = 0
∴ x = - 4    or     x = 2

Question:

Solution:
(x + 8)/x = (x + 2)/2
⇒ 2x + 16 = x² + 2x
⇒ 16 = x²
⇒ x² - 16 = 0
⇒ x² - 4² = 0
⇒ (x + 4)(x - 4) = 0
∴ x = - 4 or  x = 4

Question: 5x² - (4x - 3)(3x + 2) = ?

Solution:

Question: What is the value of the greater root of the equation x² - 5x + 4 = 0?

Solution:
x² - 5x + 4 = 0
⇒ x² - 4x - x + 4 = 0
⇒ x(x - 4) - (x - 4) = 0
⇒ (x - 1)(x - 4) = 0
so the roots of the equation are x₁ = 1 and x₂ = 4. The greater one is obviously 4.