More Equations

There are various types of equations each that have techniques which we can apply to certain equations to make them easier to solve. However, they employ the same basic algebraic rules and these rules never change.

Rational Exponents

A rational exponent is an exponent which indicates a power in the numerator and a root in the denominator. There are a various number of ways to express terms with rational exponents. This includes $a^{\frac{m}{n}} = (a^{\frac{1}{n}})^m = (a^m)^{\frac{1}{n}} = \sqrt[n]{a^m} = (\sqrt[n]{a})^m$.

A technique to solving equations with variables raised to a rational exponent is by raising both sides of the equation to the reciprocal of the exponent in order to eliminate the exponent on the variable term.

$\underline{Example}$

$x^{\frac{3}{2}} = 125$

$(x^{\frac{3}{2}})^{\frac{2}{3}} = 125^{\frac{2}{3}}$

$x = (\sqrt[3]{125})^2$

$x = 5^2$

$x = 25$

Solving By Factoring

Factoring and using the zero-product property is a valuable technique for many types of polynomial equations with degrees higher than $2$. A polynomial of degree $n$ is an expression of the type $a_nx^n + a_{n - 1}x^{n-1} + ... + a_2x^2 + a_1x + a_0$ where $n$ is a positive integer and $a_n, ... ,a_0$ are real numbers and $a_n \neq 0$. Setting the polynomial to zero converts this polynomial into a polynomial equation.

$\underline{Example}$

$12x^4 = 3x^2$

$12x^4 - 3x^2 = 0$

$3x^2(4x^2 - 1) = 0$

$4x^2 - 1 \to (2x + 1)(2x - 1)$

$3x^2(2x + 1)(2x - 1) = 0$

$3x^2 = 0 \to x = 0$

$2x + 1 = 0 \to x = -\frac{1}{2}$

$2x - 1 = 0 \to x = \frac{1}{2}$

$x = -\frac{1}{2}, 0, \frac{1}{2}$

note

The total number of solutions (real and complex) to a polynomial equation is equal or less than the highest exponent $n$.

Radical Equations

Radical equations are equations that contain terms with a variable in the radicand. These equations can be solved by eliminating each radical, one at a time. The issue with solving radical equations is that it is not unusual to find extraneous solutions which are solutions that are not valid. These solutions appear from the process of raising both sides of an equation to a power which can alter domains. The remedy to this solution is to check each answer in the original equation to confirm all true solutions.

To solve a radical equation, we isolate the radical expression on one side and the remaining terms on the other side. We then raise both sides to the $n$th power in order to elimate the root and then solve the remaining equation. We may have to repeat this process till all radicals are eliminated. Once this is done, we just need to confirm the solutions by substituting them into the original equation.

$\underline{Example}$

$\sqrt{x + 3} = 3x - 1$

$x + 3 = (3x - 1)^2$

$x + 3 = 9x^2 - 6x + 1$

$9x^2 - 7x - 2 = 0$

$9x^2 - 9x + 2x - 2 = 0$

$9x(x - 1) + 2(x - 1) = 0$

$(9x + 2)(x - 1) = 0$

$x - 1 = 0 \to x = 1$

$9x + 2 = 0 \to x = -\frac{2}{9}$

$\sqrt{1 + 3} = 3(1) - 1 \to \sqrt{4} = 3 - 1 \to 2 = 2 \:\:\:\:\:\checkmark$

$\sqrt{-\frac{2}{9} + 3} = 3(-\frac{2}{9}) - 1 \to \sqrt{\frac{25}{9}} = -\frac{2}{3} - 1 \to \frac{5}{3} = -\frac{5}{3} \:\:\:\:\:\times$

$x = 1$

Absolute Value Equations

The absolute value of $x$ is written as $|x|$ and if $x \geq 0$, then $|x| = x$. On the other hand, if $x < 0$, then $|x| = -x$. Noting this property, we can solve an absolute value equation by isolating the absolute value expression on one side of the equal sign. If $|A| = c$ where $A$ is the equation and $c > 0$, then to solve the absolute value equation, we can solve for $A = c$ and $A = -c$.

$\underline{Example}$

$|6x + 4| = 8$

$6x + 4 = 8 \to 6x = 4 \to x = \frac{2}{3}$

$6x + 4 = -8 \to 6x = -12 \to x = -2$

$x = -2, \frac{2}{3}$

note

An absolute value equation in the form $|ax + b| = c$ has no solution if $c < 0$, one solution if $c = 0$, two solutions if $c > 0$.

Quadratic Form

An equation is in quadratic form if it has three terms and the exponent in the middle term is one-half of the exponent on the leading term. The third term is a constant.

We can solve equations in quadratic form by substituting a variable such as $u$, for the variable portion of the middle term and then rewriting the equation to take on the standard form of a quadratic. Once this is accomplished, we can solve for the quadratic and then replace the subtitution variable, $u$, for the original term. Finally, we solve the remaining equation.

$\underline{Example}$

$x^4 - 8x^2 - 9 = 0$

$u = x^2$

$u^2 - 8u - 9 = 0$

$u^2 - 9u + u - 9 = 0 \to u(u - 9) + 1(u - 9) = 0 \to (u + 1)(u - 9) = 0$

$u + 1 = 0 \to x^2 + 1 = 0 \to x^2 = -1 \to x = \pm i$

$u - 9 = 0 \to x^2 = 9 \to x = \pm 3$

$x = -i, i, -3, 3$

Quadratic Form Containing Bionomials

The same method of solving equations in quadratic form can be applied even if the variable in the equation is a binomial...

$\underline{Example}$

$(x - 5)^2 - 4(x - 5) - 21 = 0$

$u = x - 5$

$u^2 - 4u - 21 = 0$

$u^2 - 7u + 3u - 21 = 0 \to u(u - 7) + 3(u - 7) = 0 \to (u + 3)(u - 7) = 0$

$u + 3 = 0 \to x - 5 + 3 = 0 \to x - 2 = 0 \to x = 2$

$u - 7 = 0 \to x - 5 - 7 = 0 \to x - 12 = 0 \to x = 12$

$x = 2, 12$