Solving Radical Equations

Solving Radical Equations exercise

• Help Leo recall how to solve his radical equations.

  Hints

  For each equation, you have to square once.

  Use opposite operations. For example:

  $\begin{array}{rcl} x+4&=&~9\\ \color{#669900}{-4} & &\color{#669900}{-4}\\ x&=&~5 \end{array}$

  Check your solutions by putting them in the original equations.

  Solution

  Let's start with the first equation Leo Vitruvian has to solve: $\sqrt x+6=9$.

  To solve $\sqrt x+6=9$ first subtract $6$ to get $\sqrt x=3$. Next square both sides of the equation to get the solution $x=9$.

  The next equation looks a little more difficult: $4=2\sqrt{\frac x3}$.

  Again we first transform the equation in such a way that we eventually have to square both sides of the equation:

  $\begin{array}{rcl} 4&=&~2\sqrt{\frac x3}\\ \color{#669900}{\div 2} & &\color{#669900}{\div 2}\\ 2 & = & ~\sqrt{\frac x3} \end{array}$

  Now we have to square to get $4=\frac x3$, and then multiply by $4$ to get $x=12$.

  The last equation is $\sqrt{5x-11}=\sqrt{x+9}$.

  Because we have a square root on both sides we first have to square both sides to get $5x-11=x+9$. This is a well known linear equation which we solve by using opposite operations:

  $\begin{array}{rcl} 5x-11&=&~x+9\\ \color{#669900}{+11} & &\color{#669900}{+11}\\ 5x&=&~x+20\\ \color{#669900}{-x} & &\color{#669900}{-x}\\ 4x&=&~20\\ \color{#669900}{\div 4} & &\color{#669900}{\div 4}\\ x&=&~5 \end{array}$

• Find the right color for $x=\sqrt{2x+8}$ by solving for $x$.

  Hints

  To check the factorization use the FOIL method for multiplying two binomials:

  F multiply the first terms.

  O multiply the outer terms.

  I multiply the inner terms.

  L multiply the last terms.

  A product equals zero if one of its factors equals zero.

  To solve the equation for each factor use opposite operations.

  Solution

  Lets solve $x=\sqrt{2x+8}$ so that we can figure out which color it belongs to.

  We have a square root on the right side and the variable $x$ on both sides.

  We still have to square first:

  $x^2=2x+8$.

  If we subtract $2x$ as well as $8$ on both sides we get a quadratic equation

  $x^2-2x-8=0$.

  Using the FOIL method we can confirm that $x^2-2x-8=(x-4)(x+2)$. so we get the following equation:

  $(x-4)(x+2)=0$.

  With the zero factor property we can conclude that either

  $\begin{array}{rcl} x-4 & = & ~0\\ \color{#669900}{+4} & &\color{#669900}{+4}\\ x& = & ~4 \end{array}$

  or

  $\begin{array}{rcl} x+2 & = & ~0\\ \color{#669900}{-2} & &\color{#669900}{-2}\\ x& = & ~-2 \end{array}$

  Be careful... if you check your solutions you get the following:

  • $4=\sqrt{2\times 4+8}=\sqrt{16}$ $~~~~~$✓
  • $-1=\sqrt{2\times (-1)+8}$ can't be true because a square root is always positive.

• Solve the radical equations.

  Hints

  Keep in mind to square once.

  You first might have to isolate the square root.

  It is always a good idea to check your solutions.

  Let's have a look at an example:

  $\sqrt x-3=3$.

  Adding $3$ leads to $\sqrt x=6$. Next we square both sides to get $x=36$.

  Checking the solution:

  $\sqrt{36}-3=6-3=3$ $~~~~~$✓

  Solution

  Keep in mind that you have to square both sides to solve radical equations. If necessary, after isolating the square root. Let's have a look at the equation we have to solve:

  $\sqrt{2x}-1=3$

  1. First we have to add $1$ to both sides to get $\sqrt{2x}=4$.
  2. Next we can square both sides to get $2x=16$.
  3. Lastly we divide by $2$ to get the solution $x=8$.

  $\sqrt{2x-1}=3$

  1. Here we can square both sides immediately to get $2x-1=9$.
  2. Now we have a linear equation we solve by using opposite operations: adding $1$ to $2x=10$ and dividing by $2$ to get the solution $x=5$.

  $2\sqrt x -1=3$

  1. Adding $1$ leads to $2\sqrt x=4$.
  2. Dividing by $2$ gives us $\sqrt x=2$.
  3. Lastly we square to get the solution $x=4$.

  $2\sqrt{x-1}=3$

  1. Dividing by $2$ leads to $\sqrt{x-1}=1.5$.
  2. Now we square to get $x-1=2.25$.
  3. Lastly we add $1$ and get the solution $x=3.25$.

• Color in the different parts of the uniform with the right colors.

  Hints

  First isolate the square root if necessary.

  If the square root is isolated, you can square both sides.

  Perhaps you have to use opposite operations to solve the resulting equation.

  Check your solutions by putting them in the origin equations.

  Solution

  Let's help Magnus find the right colors by helping her solve the radical equations she has been given:

  $~$

  $\sqrt{2x+5}-2=1$

  1. Add $2$ to get $\sqrt{2x+5}=3$.
  2. Now square both sides to get $2x+1=9$.
  3. Subtracting $1$ and dividing by $2$ after gives the solution $x=4$.

  So this part of the uniform must be colored by green.

  $~$

  $\sqrt{x+4}=\sqrt{2x-1}$

  1. Here we square both sides directly and get a linear equation $x+4=2x-1$.
  2. We solve this equation using opposite operations: subtracting $x$ and adding $1$ after to get the solution $x=5$.

  The corresponding color is blue.

  $~$

  $2\sqrt{x}-3=1$

  1. Pay attention to PEMDAS: first add $3$ to get $2\sqrt x=4$ and divide by $2$ after to get $\sqrt x=2$.
  2. Lastly square to get the solution $x=4$.

  This part of the uniform has to be colored by yellow.

  $~$

  For the last part we have to solve the equation $\sqrt{x+3}=\sqrt{4x}$.

  1. Squaring on both sides leads to the linear equation $x+3=4x$.
  2. Now subtract $x$ to get $3=3x$ and divide by $3$. The solution is $x=1$.

  This part of the uniform is violet.

• Explain how to solve radical equations.

  Hints

  The equation $x+2=2x-3$ is a linear equation.

  Subtracting $x$ and adding $3$ leads to $x=5$.

  Checking this solution, we see that

  $5+2=2\times 5-3$ $~~~~$✓

  We always use the opposite of the given operation:

  $\begin{array}{rcl} x+ 2 & = & ~2x-3\\ \color{#669900}{-x} & &\color{#669900}{-x}\\ 2 & = & ~x-3\\ \color{#669900}{+3} & & \color{#669900}{+3}\\ 5 & = & ~x \end{array}$

  The opposite operation of a square root is squaring.

  Solution

  A radical equation is characterized by a square root. The opposite operation of a square root is squaring.

  Let's have a look at an example: $\sqrt x=2$ leads to $x=4$ by squaring.

  So no matter what you have to square to get rid of a square root. But you can only do this if the square root is isolated. Otherwise you have to use opposite operations. For example:

  $\begin{array}{rcl} \sqrt x+6 & = & ~9\\ \color{#669900}{-6} & &\color{#669900}{-6}\\ \sqrt x & = & ~3 \end{array}$

  Now you can square again.

  Sometimes squaring leads to further equations, for example linear equations. You still have to use the methods you've already learned, like opposite operations, to solve these kinds of equations.

• Calculate the solutions to the given radical equations.

  Hints

  First isolate the square root, then square both sides.

  Squaring the equation $\sqrt{2x+3}=x$ leads to two solutions. If you check the two solutions, you will see that one of them actually fails.

  Solve the resulting equations using opposite operations.

  It's always better to check your solutions.

  Solution

  Let's start with $3+\sqrt x=5$. First we subtract $3$ to get $\sqrt x=2$. Squaring gives us the solution $x=4$.

  Next we have an equation with one point to take care of: $\sqrt{2x+3}=x$.

  1. Squaring both sides leads to $2x+3=x^2$.
  2. Next we divide by $2x$ as well as $3$ to get $x^2-2x-3=0$.
  3. Now we factor the left side to get $(x-3)(x+1)=0$.
  4. Using the zero factor property, we get that $x=3$ or $x=-1$.

  Remember to always check your solutions because squaring, as you can get solutions which do not satisfy the original equation:

  • $\sqrt{2\cdot 3+3}=\sqrt 9=3$ $~~~~~$✓ This is the desired solution.
  • $\sqrt{2\cdot (-1)+3}=\sqrt 1=1\neq -1$. This solution doesn't satisfy the equation.

  $~$

  Now consider $2\sqrt x+2=4$

  1. Subtract $2$ to get $2\sqrt x=2$.
  2. Divide by $2$ to get $\sqrt x=1$.
  3. Squaring gives us the solution $x=1$.

  $~$

  Finally, $\sqrt{2x-5}=\sqrt{3x-12}$

  1. Squaring first gives the linear equation $2x-5=3x-12$.
  2. Subtracting $2x$ and adding $12$ lead to the solution $x=7$.