If 2^{2x + 1} - 9(2^x) + 4 = 0, what is the product of all real values of x that satisfy t… — JAMB UTME

If $2^{2x + 1} - 9(2^x) + 4 = 0$ , what is the product of all real values of $x$ that satisfy the equation?

The product of all real values of

x

satisfying the equation is

-2

Rewriting

2^{2x+1}

2(2^x)^2

allows substitution of

u = 2^x

, giving

2u^2 - 9u + 4 = 0

. Solving for

u

gives

u = \frac{1}{2}

and

u = 4

. Converting back to

x

via

2^x = 2^{-1}

and

2^x = 2^2

yields

x = -1

and

x = 2

. The product of these roots is

(-1) \times 2 = -2

Apply the law of indices to rewrite the first term.

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

. Thus, the equation becomes

2(2^x)^2 - 9(2^x) + 4 = 0

Splitting the index using

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

enables transformation into a quadratic form.

Substitute

u = 2^x

and solve the resulting quadratic equation.

2u^2 - 9u + 4 = 0 \implies (2u - 1)(u - 4) = 0 \implies u = \frac{1}{2} \text{ or } u = 4

Algebraic substitution simplifies the exponential equation into a standard quadratic equation.

Convert the values of

u

back to

x

using index laws.

For

u = \frac{1}{2}

2^x = 2^{-1} \implies x_1 = -1

. For

u = 4

2^x = 2^2 \implies x_2 = 2

Equating bases (

2^x = 2^k \implies x = k

) isolates the unknown variable

x

Compute the product of the solutions

x_1

and

x_2

Product =

x_1 \cdot x_2 = (-1) \times 2 = -2

The question specifically requests the product of all real values of

x

Solving quadratic-form exponential equations using index laws and variable substitution.

Estimated Time:2m 0s

Question