MAT 101: More Equation Solving

Rational Equations:

The Recipe of Solving Rational Equations

  1. First find the LCD (Least Common Denominator), i.e. the Least Common Multiple of all denominators.
  2. Multiply your rational equation by the LCD.
  3. Solve the resulting equation (usually a linear equation or a quadratic equation).
  4. Plug solutions you obtained from STEP 3 into the original rational equation for $x$ to see if they all satisfy.

Example. Solve
$$\frac{1}{3x+6}-\frac{1}{x^2-4}=\frac{3}{x-2}.$$

Solution. STEP 1. $3x+6=3(x+2)$ and $x^2-4=(x+2)(x-2)$. Thus the LCD is $3(x+2)(x-2)$.

STEP 2. Multiply the equation the by LCD.
$$\frac{1}{3x+6}\cdot 3(x+2)(x-2)-\frac{1}{x^2-4}\cdot 3(x+2)(x-2)=\frac{3}{x-2}\cdot 3(x+2)(x-2)$$
which results the linear equation.
$$(x-2)-3=3\cdot 3(x+2).$$
The solution of this linear equation is $x=-\frac{23}{8}$.

Radical Equations:

The Recipe of Solving Radical Equations

  1. First isolate one radical term in one side.
  2. Square both sides of the equation.
  3. If all radical are gone, solve the resulting equation (usually linear or quadratic). If not (in case the radical equation had two radical terms), repeat the steps 1 and 2.
  4. Plug solutions you obtained from STEP 3 into the original radical equation for $x$ to see if they all satisfy.

Example. Solve $5+\sqrt{x+7}=x$.

Solution. STEP 1. Isolate the radical term $\sqrt{x+7}$ in the LHS.
$$\sqrt{x+7}=x-5.$$

STEP 2. Square both sides of the resulting equation.
$$x+7=(x-5)^2.$$
This is simplified to the quadratic equation
$$x^2-11x+18=0$$ which is factored to
$$(x-9)(x-2)=0.$$
Hence we obtain the two solutions $x=2,9$. However, not all these solutions may satisfy the original radical equation.

STEP 3. If $x=9$, then $$\mbox{LHS}=5+\sqrt{9+7}=9=\mbox{RHS}.$$ However, if $x=2$ then
$$\mbox{LHS}=5+\sqrt{x+7}=8\ne 2=\mbox{RHS}.$$ Therefore $x=9$ is the only solution to the radical equation.

Example. Solve $\sqrt{x-3}+\sqrt{x+5}=4$.

Solution. STEP 1. Isolate on radical term in one side, say isolate $\sqrt{x-3}$ in the LHS.
$$\sqrt{x-3}=4-\sqrt{x+5}.$$

STEP 2. Square both side of the resulting equation.
\begin{align*}
x-3&=(4-\sqrt{x+5})^2\\
&=16-8\sqrt{x+5}+(x+5)\\
&=x+21-8\sqrt{x+5}.
\end{align*}

Repeat STEP 1. Isolate the radical term $\sqrt{x+5}$ in the RHS.
$$3=\sqrt{x+5}.$$

Repeat STEP 2. Square both sides of the resulting equation.
$$9=x+5.$$
Hence we obtain the solution $x=4$.

One can readily check that $x=4$ satisfies the original radical equation.

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