Equations Reducible to Quadratic Form

Equations Reducible to Quadratic Form

$x = 2$ or $x=\frac{1}{2}$$x = \frac{1}{2}$ or $x = 3$ or $x=\frac{1}{3}$$x = \frac{1}{3}$

$S\mathrm{.}S\mathrm{.}=\{2,\frac{1}{2},3,\frac{1}{3}\}$$S.S. = \{2, \frac{1}{2}, 3, \frac{1}{3}\}$

Exponential Equations

An equation involving expressions of the form $k^{x}$ is called an exponential equation where $k$ is a constant. In exponential equations the exponents are variables. We will discuss only those exponential equations which can be reduced to quadratic form.

Exponential equations reducible to quadratic form are of the type:

We use the substitution $k^{x} = y$ to reduce the above equation to quadratic form. Then we find y and replace it with $k^{x}$ . In this way, roots of the exponential equations are found.

Example: Solve the following exponential equations

i. $2^{2x}-2\times 2^x+1=0$$2^{2x} - 2 \times 2^{x} + 1 = 0$

Solution:

i. $2^{2x}-2\times 2^x+1=0\to (i)$$2^{2x} - 2 \times 2^{x} + 1 = 0 \rightarrow (i)$

Put $2^{x} = y$, so $2^{2x} = y^{2}$

Equation (i) becomes:

Replace y by its substitute $2^{x}$

ii. $9^{x} - 3^{x + 1} - 4 = 0$

ii. $9^x-3^{x+1}-4=0\to (i)$$9^{x} - 3^{x + 1} - 4 = 0 \rightarrow (i)$

Put $3^{x} = y$, so $3^{2x} = y^{2}$

Equation (i) becomes: