Solve $2^{(2y+1)} - 5(2^y) + 2 = 0$ | WAEC SSCE - Further Mathematics - 2022 (Essay)

Question 1 Report

Solve $2^{(2y+1)} - 5(2^y) + 2 = 0$

Answer

$2^{(2 y + 1)} - 5 (2^{y}) + 2$ = 0

Let p = 2 $^{y}$
$2^{2 y} (2^{1}) - 5 (2^{y})$ + 2 = 0
2p $^{2}$ - 5p + 2 = 0
2p $^{2}$ - p - 4p + 2 = 0
p (2p - 1) - 2(2p - 1) = 0
(p - 2)(2p - 1) = 0

p = 2 or $\frac{1}{2}$

p = 2 $^{y}$
when p = 2
2 $^{y}$ = 2
y = 1

when p = $\frac{1}{2}$

2 $^{y}$ = $\frac{1}{2}$

2 $^{y}$ = 2 $^{- 1}$

y = -1

y = -1 or 1

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Answer Details

$2^{(2 y + 1)} - 5 (2^{y}) + 2$ = 0

Let p = 2 $^{y}$
$2^{2 y} (2^{1}) - 5 (2^{y})$ + 2 = 0
2p $^{2}$ - 5p + 2 = 0
2p $^{2}$ - p - 4p + 2 = 0
p (2p - 1) - 2(2p - 1) = 0
(p - 2)(2p - 1) = 0