Question

A)Solve the equation
`2y^2-7y-4=0`

b) Use your answer of part a to solve the equation
`2(2^x)^2-7(2^x)-4=0`

Answer 1

see explanation

Explanation:

Given

2y² - 7y - 4 = 0 ← in standard form

(2y + 1)(y - 4) = 0 ← in factored form

Equate each factor to zero and solve for y

2y + 1 = 0 ⇒ 2y = - 1 ⇒ y = -
`(1)/(2)`

y - 4 = 0 ⇒ y = 4

(b)

Comparing this equation to the one given in (a) , then

`2^(x)` = y

Using the solutions from part (a)

`2^(x)` = -
`(1)/(2)` ( multiply both sides by - 1 )

-
`2^(x)` =
`(1)/(2)` [ note that
`2^(-1)` =
`(1)/(2)` ]

x = - 1

`2^(x)` = 4 [ note that 2² = 4 ]

x = 2