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

Answer:

see explanation

Step-by-step 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 = - $\frac{1}{2}$

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)

$2^{x}$ = - $\frac{1}{2}$ ( multiply both sides by - 1 )

- $2^{x}$ = $\frac{1}{2}$ [ note that $2^{-1}$  = $\frac{1}{2}$ ]

x = - 1

$2^{x}$ = 4 [ note that 2² = 4 ]

x = 2