Question

some of roots of equation 4^x - 3(2^(x+3)) + 128 = 0 is
a)0
b)7
c)5
d)8

Answer 1

Answer :

  " Option B = 7 " is correct

Solution :

The given equation is

    4ˣ - 3(2ˣ⁺³) + 128 = 0

⇒ (2ˣ)² - 3(2ˣ × 2³) + 128 = 0

⇒ (2ˣ)² - 3(2ˣ × 8) + 128 = 0

⇒ (2ˣ)² - 24 (2ˣ) + 128 = 0

⇒ (2ˣ)² - 16 (2ˣ) - 8 (2ˣ) + 128 = 0

⇒ 2ˣ (2ˣ - 16) - 8 (2ˣ - 16) = 0

⇒ (2ˣ - 16) (2ˣ - 8) = 0

Either 2ˣ - 16 = 0 or, 2ˣ - 8 = 0

     ⇒ 2ˣ = 16 , 2ˣ = 8

     ⇒ 2ˣ = 2⁴ , 2ˣ = 2³

     ⇒ x = 4 , x = 3

The roots of the given equation are

              x = 4 , x = 3

Hence, the required sum of the roots

            = 4 + 3 = 7

Answer 2

Answer:

7

Step-by-step explanation:

4ˣ - 3(2ˣ⁺³) + 128 = 0

(2ˣ)² - 3(2ˣ × 2³) + 128 = 0

(2ˣ)² - 3(2ˣ × 8) + 128 = 0

(2ˣ)² - 24 (2ˣ) + 128 = 0

(2ˣ)² - 16 (2ˣ) - 8 (2ˣ) + 128 = 0

2ˣ (2ˣ - 16) - 8 (2ˣ - 16) = 0

(2ˣ - 16) (2ˣ - 8) = 0

2ˣ - 16 = 0 ", ˣ - 8 = 0

     2ˣ = 16 " 2ˣ = 8

    2ˣ = 2⁴ " 2ˣ = 2³

    x = 4. " x = 3

Sum of both :-

4+3=7.

Option B

Hope it will help you