If 2^{2x + 1} - 5(2^x) + 2 = 0, what are the possible values of x? — JAMB UTME

If $2^{2x + 1} - 5(2^x) + 2 = 0$ , what are the possible values of $x$ ?

-1

1

Using the addition law of indices,

2^{2x+1}

can be rewritten as

2^1 \cdot 2^{2x} = 2 \cdot (2^x)^2

. Substituting

y = 2^x

produces the quadratic equation

2y^2 - 5y + 2 = 0

, which factorizes into

(2y - 1)(y - 2) = 0

. This gives

y = \frac{1}{2}

y = 2

. Converting back to exponential equations gives

2^x = 2^{-1} \implies x = -1

and

2^x = 2^1 \implies x = 1

. Thus, the solutions for

x

are

-1

1

Apply index laws to express the equation in terms of

2^x

2^{2x+1} = 2^{2x} \cdot 2^1 = 2 \cdot (2^x)^2

, so the equation becomes

2 \cdot (2^x)^2 - 5(2^x) + 2 = 0

Splitting the exponent using

a^{m+n} = a^m \cdot a^n

reveals a quadratic structure in

2^x

Substitute

y = 2^x

and solve the resulting quadratic equation

2y^2 - 5y + 2 = 0 \implies (2y - 1)(y - 2) = 0 \implies y = \frac{1}{2} \text{ or } y = 2

Factoring the quadratic expression yields the values for the substitution variable

y

Equate

2^x

to each solution of

y

to solve for

x

For

y = \frac{1}{2}

2^x = 2^{-1} \implies x = -1

. For

y = 2

2^x = 2^1 \implies x = 1

Using the negative index rule

a^{-n} = \frac{1}{a^n}

allows matching exponents when bases are identical.

Quadratic Equations Reducible to Index Form

Soru