4.06 Solve radical equations

\sqrt{y}+5=9

To solve for y, we first want to isolate \sqrt{y}. We can then square both sides of the equation, solve the resulting equation, and check our solution(s).

If we substitute y=16 into the original equation, we obtain \sqrt{16}+5=9 which is a true statement, so the solution is valid. Remember, we must always check our solution(s) for possible extraneous solutions.

\sqrt[3]{4x-9}=-1

The cube root expression is already isolated on one side of equation. To solve for x, we will raise each side of the equation to a power of 3.

If we substitute x=2 into the original equation, we obtain \sqrt[3]{-1}=-1 which is a true statement. When dealing with odd valued indexes such as \sqrt[3]{x}, we do not need to worry about the radicand being negative.

Raising both sides of an equation to an odd power is reversible. If a=b, then a^3=b^3 and if a^3=b^3, then a=b.

3x=1+2\sqrt{x}

To solve for x, we first want to isolate 2\sqrt{x}. We can then square both sides of the equation to remove the radical. We can then solve the resulting quadratic equation. We then want to check for extraneous solutions.

1. \displaystyle 3x	\displaystyle =	\displaystyle 1+2\sqrt{x}	State the equation
   \displaystyle 3x-1	\displaystyle =	\displaystyle 2\sqrt{x}	Subtract 1 from both sides
   \displaystyle \left(3x-1\right)^2	\displaystyle =	\displaystyle \left(2\sqrt{x}\right)^2	Square both sides
   \displaystyle 9x^2-6x+1	\displaystyle =	\displaystyle 4x	Evaluate the exponents
   \displaystyle 9x^2-10x+1	\displaystyle =	\displaystyle 0	Subtract 4x from both sides
   \displaystyle \left(9x-1\right)\left(x-1\right)	\displaystyle =	\displaystyle 0	Factor the quadratic

   Using the zero product property we get two solutions x=\dfrac{1}{9}, x=1. We now want to test both solutions:

   \displaystyle 3\left(1\right)	\displaystyle =	\displaystyle 1+2\sqrt{1}	Substitute x=1
   \displaystyle 3	\displaystyle =	\displaystyle 3	Evaluate the multiplication and addition

   We can see that x=1 satisfies the original equation and is a valid solution.

   Now, let's test x=\dfrac{1}{9}.

   \displaystyle 3\left(\dfrac{1}{9}\right)	\displaystyle =	\displaystyle 1+2\sqrt{\frac{1}{9}}	Substitute x=\dfrac{1}{9}
   \displaystyle \frac{1}{3}	\displaystyle =	\displaystyle \frac{5}{3}	Evaluate the multiplication and addition

   We can see that x=\dfrac{1}{9} leads to a false statement and is therefore an extraneous solution.

2. The equation 3x=1+2\sqrt{x} can be written as the system of equations: \begin{cases} y= 3x \\ y= 1 + 2 \sqrt{x} \end{cases}

3. We can use technology to graph the system of equations.

   

We can see the solution x=1 as the point of intersection on the graph, while the extraneous solution found algebraically, x=\dfrac{1}{9}, does not appear on the graph.

\sqrt{x+17}=x+5

To solve for x, we first want to square both sides of the equation to remove the radical. We can then solve the resulting quadratic equation. We then want to check for extraneous solutions.

1. \displaystyle \sqrt{x+17}	\displaystyle =	\displaystyle x+5	State the equation
   \displaystyle \left(\sqrt{x+17}\right)^2	\displaystyle =	\displaystyle \left(x+5\right)^2	Square both sides
   \displaystyle x+17	\displaystyle =	\displaystyle x^2+10x+25	Evaluate the exponents
   \displaystyle 0	\displaystyle =	\displaystyle x^2+9x+8	Subtract x and 17 from both sides
   \displaystyle 0	\displaystyle =	\displaystyle \left(x+1\right)\left(x+8\right)	Factor the quadratic

   Using the zero product property we get two solutions x=-8, x=-1. We now want to test both solutions:

   \displaystyle \sqrt{-8+17}	\displaystyle =	\displaystyle -8+5	Substitute x=-8
   \displaystyle 3	\displaystyle =	\displaystyle -3	Evaluate the addition and square root

   We can see that x=-8 leads to a false statement and is therefore an extraneous solution.

   Now, let's test x=-1.

   \displaystyle \sqrt{-1+17}	\displaystyle =	\displaystyle -1+5	Substitute x=-1
   \displaystyle 4	\displaystyle =	\displaystyle 4	Evaluate the addition and square root

   We can see that x=-1 satisfies the original equation and is a valid solution.

2. The equation\sqrt{x+17}=x+5 can be written as the system of equations: \begin{cases} y= \sqrt{x+17} \\ y= x+5 \end{cases}

3. We can use technology to graph the system of equations.

   

It is important to note that the fact that our solutions were negative, x=-8, x=-1, does not necessarily make them extraneous. We can see after substitution that only x=- 8 is an extraneous solution and not a point of intersection on the graph of the system of equations, and x=-1 is valid.

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

To solve for x, we first want to square both sides of the equation to remove the radical. We can then solve the resulting quadratic equation. We then want to check for extraneous solutions.

1. \displaystyle 4\sqrt{x-1}	\displaystyle =	\displaystyle x+2	State the equation
   \displaystyle \left(4\sqrt{x-1}\right)^2	\displaystyle =	\displaystyle \left(x+2\right)^2	Square both sides
   \displaystyle 16(x-1)	\displaystyle =	\displaystyle x^2+4x+4	Evaluate the exponents
   \displaystyle 16x-16	\displaystyle =	\displaystyle x^2+4x+4	Distributive property
   \displaystyle 0	\displaystyle =	\displaystyle x^2-12x+20	Subtract 16x from and add 16 to both sides
   \displaystyle 0	\displaystyle =	\displaystyle \left(x-10\right)\left(x-2\right)	Factor the quadratic

   Using the zero product property, we get two solutions: x=10, x=2. We now want to test both solutions:

   \displaystyle 4\sqrt{10-1}	\displaystyle =	\displaystyle 10+2	Substitute x=10
   \displaystyle 12	\displaystyle =	\displaystyle 12	Evaluate the addition and square root

   We can see that x=10 leads to a true statement and is therefore a valid solution.

   Now, let's test x=2.

   \displaystyle 4\sqrt{2-1}	\displaystyle =	\displaystyle 2+2	Substitute x=2
   \displaystyle 4	\displaystyle =	\displaystyle 4	Evaluate the addition and square root

   We can see that x=2 satisfies the original equation and is a valid solution.

2. The equation4\sqrt{x-1}=x+2 can be written as the system of equations: \begin{cases} y= 4\sqrt{x-1} \\ y= x+2 \end{cases}

3. We can use technology to graph the system of equations.

   

Notice that we did not solve the equation by isolating the radical first. When the radical has a coefficient, it is not necessary to isolate it before solving. In this case, isolating the radical first requires a couple more steps.

Following the same steps as before, we arrive at the same solutions for x, which are x=10 and x=2. This confirms that our original solution is correct, and the alternate method of isolating the radical first yields the same result.