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## Find the roots of basic equations containing roots

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Can you solve the following equation?

\begin{align*}x+ \sqrt{x-2}=4\end{align*}

A radical equation is an equation with a variable under a radical sign. The following are all examples of radical equations:

• \begin{align*}\sqrt{x}=5\end{align*}
• \begin{align*}\sqrt{x-4}+5=0\end{align*}
• \begin{align*}x+\sqrt{x-2}=4\end{align*}
• \begin{align*}\sqrt{4x+5}-\sqrt{2x-6}=3\end{align*}

Just like multiplication and division or addition and subtraction are inverse operations (they "undo" each other), squaring and finding a square root are inverse operations:

• \begin{align*}(\sqrt{5})^2=5\end{align*}
• \begin{align*}\sqrt{(x+2)^2}=x+2\end{align*}

Therefore, to eliminate a square root in an equation, isolate the square root part of the equation and then square both sides of the equation. If there are multiple square roots, you might have to go through this process multiple times. For example, consider the following equation:

\begin{align*}\sqrt{x-1}-5=0\end{align*}

\begin{align*}\sqrt{x-1}=5\end{align*}

Step 2: Square both sides of the equation.

\begin{align*}& \left(\sqrt{x-1}\right)^{{\color{red}{2}}}=(5)^{\color{red}{2}} \\ & x-1=25\end{align*}

Step 3: Solve the resulting equation.

\begin{align*}x=26\end{align*}

Just like any equation, you can check your answer(s) by substituting them back into the original equation. Unlike other equations, it is very important to check your answers to radical equations. Due to the fact that the process of squaring produces positive numbers, sometimes you will end up with a solution to a radical equation that does not actually work in the original equation. These solutions are called "extraneous" and are not actually solutions. Therefore, you must always check your answers to radical equations. Check the solution to the above equation by verifying that the left side (L.S.) of the equation equals the right side (R.S.) of the equation:

\begin{align*}&\sqrt{x-1}-5=0 \\ & L.S.= \sqrt{x-1}-5 \qquad R.S.=0 \\ & L.S.= \sqrt{{\color{red}{26}}-1}-5 \\ & L.S.= \sqrt{25}-5 \\ & L.S.= {\color{red}{5}}-5 \\ & L.S.=0 \\ & L.S.=R.S.\end{align*}

Therefore, the solution of 26 is correct.

#### Let's practice solving radical equations and verifying the solution(s) for the following equations:

1. \begin{align*}2\sqrt{x-1}-1=9\end{align*}

\begin{align*}2\sqrt{x-1}-1{\color{red}{+1}}&=9{\color{red}{+1}} \\ 2\sqrt{x-1}&=10 \\ \frac{2\sqrt{x-1}}{{\color{red}{2}}}&= \frac{10}{{\color{red}{2}}} \\ \sqrt{x-1}&=5\end{align*}

Now square both sides of the equation:

\begin{align*}\left(\sqrt{x-1}\right)^{\color{red}{2}}&= (5)^{\color{red}{2}} \\ x-1&=25\end{align*}

Solve the equation:

\begin{align*}x&=26\end{align*}

Finally, verify the result by substituting the value of 26 for ‘\begin{align*}x\end{align*}’ into the original equation. If 26 is a solution to the equation, the left side (L.S.) will equal the right side (R.S.).

\begin{align*}&2\sqrt{x-1}-1=9 \\ & L.S.=2\sqrt{{\color{red}{26}}-1}-1 \quad R.S.=9 \\ & L.S.=2\sqrt{{\color{red}{25}}}-1 \\ & L.S.=2({\color{red}{5}})-1 \\ & L.S.={\color{red}{10}}-1 \\ & L.S.=9 \\ & L.S.=R.S. \end{align*}

Therefore, the solution of 26 is correct.

1. \begin{align*}\sqrt{4x+5}- \sqrt{2x-6}=3\end{align*}

When a radical equation has more than one radical, begin by writing the equation with one radical on each side of the equation. You will ultimately have to square the equation more than once throughout the solution process.

\begin{align*}&\sqrt{4x+5}=3+ \sqrt{2x-6}\end{align*}

Square both sides of the equation:

\begin{align*}\left(\sqrt{4x+5}\right)^{\color{red}{2}}=\left(3+ \sqrt{2x-6}\right)^{\color{red}{2}}\end{align*}

Expand and simplify:

\begin{align*}&\left(\sqrt{4x+5}\right)^{\color{red}{2}}=\left(3+ \sqrt{2x-6}\right)\left(3+ \sqrt{2x-6}\right) \\ &4x+5=9+3 \sqrt{2x-6}+3 \sqrt{2x-6}+2x-6 \\ &4x+5=3+6 \sqrt{2x-6}+2x\end{align*}

\begin{align*}&2x+2=6 \sqrt{2x-6}\end{align*}

Simplify the equation:

\begin{align*}&\frac{2x}{{\color{red}{2}}}+ \frac{2}{{\color{red}{2}}}= \frac{6 \sqrt{2x-6}}{{\color{red}{2}}} \\ & x+1=3 \sqrt{2x-6}\end{align*}

Square both sides of the equation:

\begin{align*}&(x+1)^{\color{red}{2}}=\left(3 \sqrt{2x-6}\right)^{\color{red}{2}} \\ & x^2+x+x+1=9(2x-6) \\ & x^2+2x+1=18x-54\end{align*}

The equation is quadratic. Write the equation in standard form:

\begin{align*}& x^2-16x+55=0\end{align*}

Solve the equation by factoring:

\begin{align*}&(x-11)(x-5)=0 \\ & x={\color{red}{11}} \ or \ x={\color{red}{5}}\end{align*}

Verify the results by first substituting the value of 11 for ‘\begin{align*}x\end{align*}’ into the original equation and then substituting the value of 5 for ‘\begin{align*}x\end{align*}’ into the original equation:

\begin{align*}&\text{Verify} \ x=11. && \text{Verify} \ x=5. \\ &\sqrt{4x+5}- \sqrt{2x-6}=3 && \sqrt{4x+5}- \sqrt{2x-6}=3 \\ & L.S.=\sqrt{4x+5}- \sqrt{2x-6} \quad R.S.=3 && L.S.=\sqrt{4x+5}- \sqrt{2x-6} \quad R.S.=3 \\ & L.S.=\sqrt{4 \left({\color{red}{11}}\right)+5}- \sqrt{2 \left({\color{red}{11}}\right)-6} && L.S.=\sqrt{4 \left({\color{red}{5}}\right)+5}- \sqrt{2 \left({\color{red}{5}}\right)-6} \\ & L.S.=\sqrt{{\color{red}{49}}}- \sqrt{{\color{red}{16}}} && L.S.=\sqrt{{\color{red}{25}}}- \sqrt{{\color{red}{4}}} \\ & L.S.={\color{red}{7}}-{\color{red}{4}} && L.S.={\color{red}{5}}-{\color{red}{2}} \\ & L.S.=3 && L.S.=3 \\ & R.S.=3 && R.S.=3 \\ & L.S.=R.S. && L.S.=R.S. \end{align*}

Therefore, the solutions of 11 and 5 are both correct.

1. \begin{align*}\sqrt{5x+6}-4=0\end{align*}

\begin{align*}\sqrt{5x+6}&=4\end{align*}

Square both sides of the equation:

\begin{align*}\left(\sqrt{5x+6}\right)^{\color{red}{2}}&=(4)^{\color{red}{2}} \\ 5x+6&=16\end{align*}

Solve the equation:

\begin{align*}5x&=10 \\ \frac{5x}{{\color{red}{5}}}&= \frac{10}{{\color{red}{5}}} \\ x&=2\end{align*}

Verify the result by substituting the value of 2 for ‘\begin{align*}x\end{align*}’ into the original equation:

\begin{align*}&\text{Verify} \ x=2. \\ &\sqrt{5x+6}-4=0 \\ & L.S.= \sqrt{5x+6}-4 \quad R.S.=0 \\ & L.S.= \sqrt{5 \left({\color{red}{2}}\right)+6}-4 \\ & L.S.= \sqrt{{\color{red}{10}}+6}-4 \\ & L.S.= \sqrt{{\color{red}{16}}}-4 \\ & L.S.={\color{red}{4}}-4 \\ & L.S.=0 \\ & R.S.=0 \\ & L.S.=R.S. \end{align*}

Therefore, the solution of 2 is correct.

### Examples

#### Example 1

Earlier, you were asked if you could solve the following equation:

\begin{align*}x+ \sqrt{x-2}=4\end{align*}

\begin{align*}&\sqrt{x-2}=4-x\end{align*}

Square both sides of the equation:

\begin{align*}\left(\sqrt{x-2}\right)^{\color{red}{2}}=(4-x)^{\color{red}{2}}\end{align*}

Expand and simplify:

\begin{align*}& x-2=16-8x+x^2\end{align*}

The equation is a quadratic. Write the equation in standard form:

\begin{align*}& x^2-9x+18=0\end{align*}

Solve the equation:

\begin{align*}&(x-6)(x-3)=0 \\ & x-6=0 \ or \ x-3=0 \\ & x={\color{red}{6}} \ or \ x={\color{red}{3}}\end{align*}

Verify the results by first substituting the value of 6 for ‘\begin{align*}x\end{align*}’ into the original equation and then substituting the value of 3 for ‘\begin{align*}x\end{align*}’ into the original equation.

\begin{align*}& \text{Verify} \ x=6. && \text{Verify} \ x=3. \\ & x+ \sqrt{x-2}=4 && x+ \sqrt{x-2}=4 \\ & L.S.={\color{red}{6}}+ \sqrt{{\color{red}{6}}-2} \quad R.S.=4 && L.S.={\color{red}{3}}+ \sqrt{{\color{red}{3}}-2} \quad R.S.=4 \\ & L.S.=6+ \sqrt{{\color{red}{4}}} && L.S.=3+ \sqrt{{\color{red}{1}}} \\ & L.S.=6+{\color{red}{2}} && L.S.=3+{\color{red}{1}}=4 \\ & L.S.=8 && L.S.=4 \\ & R.S.=4 && R.S.=4 \\ & L.S.\ne R.S. && L.S.=R.S. \end{align*}

The value \begin{align*}x=6\end{align*} did not satisfy the original equation and is not a solution to the radical equation. It is called an extraneous solution. \begin{align*}x=3\end{align*} is a solution to the equation.

#### Example 2

Verify whether or not \begin{align*}x=3\end{align*} and \begin{align*}x=-5\end{align*} are solutions to the radical equation:

\begin{align*}2\sqrt{x+6}+ \sqrt{2x+10}=2\end{align*}

Verify the results by first substituting the value of 3 for ‘\begin{align*}x\end{align*}’ into the original equation and then substituting the value of -5 for ‘\begin{align*}x\end{align*}’ into the original equation.

\begin{align*}&\text{Verify} \ x=3. && \text{Verify} \ x=-5. \\ & 2 \sqrt{x+6}+ \sqrt{2x+10}=2 && 2 \sqrt{x+6}+ \sqrt{2x+10}=2 \\ & L.S.=2 \sqrt{x+6}+ \sqrt{2x+10} \quad R.S.=2 && L.S.=2\sqrt{x+6}+ \sqrt{2x+10} \quad R.S.=2 \\ & L.S.=2 \sqrt{\left({\color{red}{3}}\right)+6}+ \sqrt{2\left({\color{red}{3}}\right)+10} && L.S.=2 \sqrt{\left({\color{red}{-5}}\right)+6}+ \sqrt{2\left({\color{red}{-5}}\right)+10} \\ & L.S.=2 \sqrt{{\color{red}{9}}}+ \sqrt{{\color{red}{16}}} && L.S.=2({\color{red}{1}})+ \sqrt{{\color{red}{0}}} \\ & L.S.=2({\color{red}{3}})+{\color{red}{4}} && L.S.=2({\color{red}{1}})+{\color{red}{0}} \\ & L.S.={\color{red}{6+4}} && L.S.={\color{red}{2+0}} \\ & L.S.=10 && L.S.=2 \\ & R.S.=2 && R.S.=2 \\ & L.S. \ne R.S. && L.S.=R.S. \end{align*}

The value \begin{align*}x=3\end{align*} did not satisfy the original equation and is not a solution to the radical equation. It is an extraneous solution. \begin{align*}x=-5\end{align*} is a solution.

#### Example 3

Solve the following radical equation and verify the solution(s) to the equation.

\begin{align*}x- \sqrt{x-1}=7\end{align*}

\begin{align*}- \sqrt{x-1}=7-x\end{align*}

\begin{align*}x-1&=49-14x+x^2\end{align*}

\begin{align*}x^2-15x+50&=0\end{align*}

\begin{align*}&(x-10)(x-5)=0 \\ & x={\color{red}{10}} \ or \ x={\color{red}{5}}\end{align*}

\begin{align*}& \text{Verify} \ x=10. && \text{Verify} \ x=5. \\ & x- \sqrt{x-1}=7 && x- \sqrt{x-1}=7 \\ & L.S.=x- \sqrt{x-1} \quad R.S.=7 && L.S.=x- \sqrt{x-1} \quad R.S.=7 \\ & L.S.=({\color{red}{10}})- \sqrt{\left({\color{red}{10}}\right)-1} && L.S.=({\color{red}{5}})- \sqrt{\left({\color{red}{5}}\right)-1} \\ & L.S.={\color{red}{10}}- \sqrt{{\color{red}{9}}} && L.S.={\color{red}{5}}- \sqrt{{\color{red}{4}}} \\ & L.S.={\color{red}{10}}-{\color{red}{3}} && L.S.={\color{red}{5}}-{\color{red}{2}} \\ & L.S.=7 && L.S.=3 \\ & R.S.=7 && R.S.=7 \\ & L.S.=R.S. && L.S. \ne R.S. \end{align*}

The value \begin{align*}x=5\end{align*} did not satisfy the original equation and is not a solution to the radical equation. It is an extraneous solution. \begin{align*}x=10\end{align*} is a solution.

#### Example 4

Solve the following radical equation and verify the solution(s) to the equation.

\begin{align*}\sqrt{x+7}- \sqrt{x}=1\end{align*}

\begin{align*}\sqrt{x+7}- \sqrt{x}=1\end{align*}

\begin{align*}\sqrt{x+7}&=1+ \sqrt{x}\end{align*}

\begin{align*}\left(\sqrt{x+7}\right)^{\color{red}{2}}= \left(1+ \sqrt{x}\right)^{\color{red}{2}}\end{align*}

\begin{align*}x+7&=1+2 \sqrt{x}+x\end{align*}

\begin{align*}6&=2 \sqrt{x}\end{align*}

\begin{align*}(6)^{\color{red}{2}}&= \left(2 \sqrt{x}\right)^{\color{red}{2}} \\ 36&=4x\end{align*}

\begin{align*}\frac{36}{{\color{red}{4}}}&= \frac{4x}{{\color{red}{4}}} \\ {\color{red}{9}}&=x\end{align*}

\begin{align*}&\text{Verify} \ x=9. \\ &\sqrt{x+7}- \sqrt{x}=1 \\ & L.S.= \sqrt{x+7}- \sqrt{x} \quad R.S.=1 \\ & L.S.= \sqrt{\left({\color{red}{9}}\right)+7}- \sqrt{{\color{red}{9}}} \\ & L.S.= \sqrt{{\color{red}{16}}}- \sqrt{{\color{red}{9}}} \\ & L.S.={\color{red}{4-3}} \\ & L.S.={\color{red}{1}} \\ & L.S.=1 \\ & R.S.=1 \\ & L.S.=R.S. \end{align*}

\begin{align*}x=9\end{align*} is a solution to the equation.

### Review

1. Is \begin{align*}x=7\end{align*} a solution to \begin{align*}\sqrt{x+2}=-3\end{align*}?
2. Is \begin{align*}x=1\end{align*} a solution to \begin{align*}\sqrt{x^2+4x+4}-\sqrt{x^2+3x}=1\end{align*}?
3. Is \begin{align*}x=-3\end{align*} a solution to \begin{align*}\sqrt{x^2+4x+4}-\sqrt{x^2+3x}=1\end{align*}?
4. Is \begin{align*}x=12\end{align*} a solution to \begin{align*}\sqrt{x+4}+8=x\end{align*}?
5. Is \begin{align*}x=5\end{align*} a solution to \begin{align*}\sqrt{x+4}+8=x\end{align*}?
6. Is \begin{align*}x=8\end{align*} a solution to \begin{align*}\sqrt{x+1}=1+ \sqrt{x-4}\end{align*}?
7. Is \begin{align*}x=3\end{align*} a solution to \begin{align*}\sqrt{x+1}+ \frac{2}{\sqrt{x+1}}= \sqrt{x+6}\end{align*}?

Solve the following radical equations and verify the solution(s).

1. \begin{align*}x=3+ \sqrt{x-1}\end{align*}
2. \begin{align*}\sqrt{x+1}=1+ \sqrt{x-4}\end{align*}
3. \begin{align*}\sqrt{x}- \sqrt{x-16}=2\end{align*}
4. \begin{align*}\sqrt{3x-2}-1= \sqrt{2x-3}\end{align*}
5. \begin{align*}5 \sqrt{x-6}=x\end{align*}
6. \begin{align*}x=5+ \sqrt{x-4}\end{align*}
7. \begin{align*}\sqrt{x+2}=5- \sqrt{x-3}\end{align*}
8. \begin{align*}\sqrt{x}+ \sqrt{x-9}=5\end{align*}

To see the Review answers, open this PDF file and look for section 9.9.

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### Vocabulary Language: English

Extraneous Solution

An extraneous solution is a solution of a simplified version of an original equation that, when checked in the original equation, is not actually a solution.

A quadratic equation is an equation that can be written in the form $=ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are real constants and $a\ne 0$.

The quadratic formula states that for any quadratic equation in the form $ax^2+bx+c=0$, $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$.