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When working with some topics such as kinetic energy or the free-fall of an object, *radical equations and inequalities* can be used to find desired values. This lesson will cover this type of equation and inequality in detail and to solve them. ### Catch-Up and Review

**Here are a few recommended readings before getting started with this lesson.**

LaShay and Ali are members of the Jefferson High School Mountain Bike Team. They are both training hard for the upcoming Mountain Bike Race.

The following equations represent their paces.$LaShay:Ali: d_{1}=10t+2t d_{2}=102t $

In these equations, $d_{1}$ and $d_{2}$ represent the distance traveled by each person, in miles, after $t$ hours. After how many hours will they both cycle the same distance? {"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":"hours","answer":{"text":["2"]}}

A radical equation is an equation that has an independent variable in the radicand of a radical expression, or an independent variable with a rational exponent. A radical equation with a root that has an index of $2$ is called a square root equation. If the index of the root is $3,$ the equation is a cube root equation.

Variable in a Radicand | Variable With a Rational Exponent |
---|---|

$x−2 =x+2$ | $x_{21}=3x−5$ |

$32x =x+1$ | $1−3x=(x−1)_{31}$ |

$3=42x+1 $ | $(2x)_{41}=4x−9$ |

Radical equations can be solved algebraically by using inverse operations and the Properties of Equality. Because some roots can only take certain values, this process can produce solutions that do not actually satisfy the equation. This type of solutions have a special name.

Solving equations by using inverse operations may lead to solutions that do not satisfy the original equation. These solutions are called extraneous solutions, and they occur primarily in radical equations when the radical is eliminated. Below, an example radical equation is shown.

For the example equation, $x=1$ is a valid solution but $x=5$ does not result in a true statement. Therefore, it is an extraneous solution.

$2=x+2x−1 $

By using inverse operations and Properties of Equality, the solutions to this equation are found to be $x=1$ and $x=5.$ However, they produce different outcomes when substituted into the original equation. Substitute | Simplify | |
---|---|---|

$x=1$ | $2=?1+2(1)−1 $ | $2=2$ $✓$ |

$x=5$ | $2=?5+2(5)−1 $ | $2 =8$ $×$ |

Consider an example radical equation.
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$2=x+2x−1 $

Four steps must be followed to solve this equation.
1

Isolate the Radical

When solving a radical equation, it is necessary to isolate the radical expression on one side. Use inverse operations to achieve this.

$2=x+2x−1 ⇕2−x=2x−1 $

2

Eliminate the Radical

The radical can be undone or eliminated by raising it to the same power as its index. Here, the radical is a square root, so the index is $2.$ Therefore, both sides of the equation must be raised to the power of $2.$

3

Solve the Equation

Once the radical has been eliminated, the resulting equation can be solved.
Now the equation can be solved for the variable. In this case, a quadratic equation was obtained and must be solved as such. First, it will be rearranged, then the Quadratic Formula will be used.
The solutions to the equation are $x=5$ and $x=1.$

$(2−x)_{2}=2x−1$

ExpandNegPerfectSquare

$(a−b)_{2}=a_{2}−2ab+b_{2}$

$2_{2}−2(2)x+x_{2}=2x−1$

CalcPowProd

Calculate power and product

$4−4x+x_{2}=2x−1$

$4−4x+x_{2}=2x−1$

SubEqn

$LHS−2x=RHS−2x$

$4−6x+x_{2}=-1$

AddEqn

$LHS+1=RHS+1$

$5−6x+x_{2}=0$

CommutativePropAdd

Commutative Property of Addition

$x_{2}−6x+5=0$

Solve using the quadratic formula

UseQuadForm

Use the Quadratic Formula: $a=1,b=-6,c=5$

$x=2(1)-(-6)±(-6)_{2}−4(1)(5) $

NegNeg

$-(-a)=a$

$x=2(1)6±(-6)_{2}−4(1)(5) $

CalcPowProd

Calculate power and product

$x=26±36−20 $

SubTerm

Subtract term

$x=26±16 $

CalcRoot

Calculate root

$x=26±4 $

StateSol

State solutions

$x=26+4 x=26−4 (I)(II) $

$(I), (II):$ Add and subtract terms

$x=210 x=22 $

$(I), (II):$ Calculate quotient

$x=5x=1 $

4

Check for Extraneous Solutions

The solutions found in the previous step might be extraneous solutions. Therefore, they must be verified in the original equation. First, the solution $x=5$ is tested.
Since $x=5$ does not satisfy the radical equation, it is an extraneous solution. Next, $x=1$ can be checked in the same way.
Since $x=1$ makes a true statement, it is a solution to the radical equation. The given equation can be said to have one solution and one extraneous solution.

$2=x+2x−1 $

Substitute

$x=1$

$2=?1+2(1)−1 $

Evaluate right-hand side

IdPropMult

Identity Property of Multiplication

$2=?1+2−1 $

SubTerm

Subtract term

$2=?1+1 $

CalcRoot

Calculate root

$2=?1+1$

AddTerms

Add terms

$2=2✓$

The rectangular board in LaShay's class is $x+5 $ wide feet by $8$ feet long.

Knowing that the perimeter of the board is $28$ feet, LaShay wants to write an equation that represents the perimeter.{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["x"],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"Equation:","formTextAfter":null,"answer":{"text":["2\\sqrt{x+5}+16=28","2(\\sqrt{x+5}+8)=28","16+2\\sqrt{x+5}=28","2(+8\\sqrt{x+5})=28"]}}

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.43056em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\">x<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["31"]}}

The perimeter of a rectangle is $P=2ℓ+2w,$ where $ℓ$ is the length and $w$ is the width of the rectangle.

The perimeter of a rectangle is $P=2ℓ+2w,$ where $ℓ$ is the length and $w$ is the width of the rectangle. A radical equation representing the perimeter of the board can be written by substituting $P=28,$ $ℓ=8,$ and $w=x+5 $ into the formula.
Now that the radical expression is isolated, the radical equation can be solved by eliminating the radical.
Finally, $x=31$ will be substituted into the original equation to see whether it is an extraneous solution.
Since it satisfies the original equation, $x=31$ is a valid solution.

$P=2ℓ+2w⇓28=2(8)+2x+5 $

From here, the first step to solving this equation is to isolate the radical expression.
$28=2(8)+2(x+5 )$

Multiply

Multiply

$28=16+2x+5 $

SubEqn

$LHS−16=RHS−16$

$12=2x+5 $

DivEqn

$LHS/2=RHS/2$

$6=x+5 $

RearrangeEqn

Rearrange equation

$x+5 =6$

$x+5 =6$

RaiseEqn

$LHS_{2}=RHS_{2}$

$(x+5 )_{2}=36$

PowSqrt

$(a )_{2}=a$

$x+5=36$

SubEqn

$LHS−5=RHS−5$

$x=31$

After an exhausting school day, LaShay went to an amusement park to have fun with her friends. The first thing they did was ride on a Ferris wheel.

Everything was going well until LaShay accidentally dropped her phone from the top of the Ferris wheel. The following radical equation models the time passed after the phone was dropped.$t=41 d−h +1 $

Here, $t$ is the time in seconds after the phone was dropped, $h$ is the height of the phone above the ground in meters, and $d$ is LaShay's height in meters above the ground at the moment she dropped the phone. If LaShay was $100$ meters high when she dropped the phone, how many meters above the ground would the phone be after $3$ seconds? {"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":"meters","answer":{"text":["36"]}}

Substitute the given values into the equation and solve for $h.$

To find how many meters above the ground the phone would be after $3$ seconds, begin by substituting $t=3$ and $d=100$ into the formula.
Finally, check if the solution satisfies the original equation.
Therefore, LaShay's phone would be $36$ meters above the ground $3$ seconds after she dropped it. Sadly, it is very unlikely that the phone will survive the fall.

$t=41 d−h +1⇓3=41 100−h +1 $

Next, the radical equation can be solved algebraically to find the value of $h.$
$3=41 100−h +1$

SubEqn

$LHS−1=RHS−1$

$2=41 100−h $

MultEqn

$LHS⋅4=RHS⋅4$

$8=100−h $

RaiseEqn

$LHS_{2}=RHS_{2}$

$64=(100−h )_{2}$

PowSqrt

$(a )_{2}=a$

$64=100−h$

AddEqn

$LHS+h=RHS+h$

$64+h=100$

SubEqn

$LHS−64=RHS−64$

$h=36$

$3=41 100−h +1$

Substitute

$h=36$

$3=?41 100−36 +1$

Evaluate right-hand side

SubTerm

Subtract term

$3=?41 64 +1$

CalcRoot

Calculate root

$3=?41 (8)+1$

MoveRightFacToNumOne

$b1 ⋅a=ba $

$3=?48 +1$

CalcQuot

Calculate quotient

$3=?2+1$

\AddTerm

$3=3✓$

After an exciting day at the amusement park, even if the math did not help LaShay save her phone from smashing against the ground, her interest in radical equations increased. Right now, LaShay and Ali are trying to solve the following radical equation.

After some work, they managed to solve the equation. However, they came up with different solutions.

Solve the given equation and decide who is correct. If LaShay's solution is wrong, she will need to study more.{"type":"choice","form":{"alts":["Ali","LaShay","Both","Neither"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":0}

Start by isolating the radical expression.

When solving a radical equation, the first thing to do is isolate the radical expression.
Since the obtained equation is a quadratic equation, the Quadratic Formula will be used to solve it.
Finally, check for extraneous solutions.

$x+-x+4 =2⇕-x+4 =2−x $

Then, raise both sides of the equation to the second power in order to get rid of the root.
$-x+14 =2−x$

RaiseEqn

$LHS_{2}=RHS_{2}$

$(-x+14 )_{2}=(2−x)_{2}$

PowSqrt

$(a )_{2}=a$

$-x+14=(2−x)_{2}$

ExpandNegPerfectSquare

$(a−b)_{2}=a_{2}−2ab+b_{2}$

$-x+14=4−4x+x_{2}$

Rewrite

AddEqn

$LHS+x=RHS+x$

$14=4−3x+x_{2}$

SubEqn

$LHS−14=RHS−14$

$0=-10−3x+x_{2}$

CommutativePropAdd

Commutative Property of Addition

$0=x_{2}−3x−10$

RearrangeEqn

Rearrange equation

$x_{2}−3x−10=0$

UseQuadForm

Use the Quadratic Formula: $a=1,b=-3,c=-10$

$x=2(1)-(-3)±(-3)_{2}−4(1)(-10) $

Evaluate right-hand side

NegNeg

$-(-a)=a$

$x=2(1)3±(-3)_{2}−4(1)(-10) $

CalcPowProd

Calculate power and product

$x=23±9+40 $

AddTerms

Add terms

$x=23±49 $

CalcRoot

Calculate root

$x=23±7 $

StateSol

State solutions

$∣∣ x=23+7 ∣∣ x=23−7 (I)(II) $

$(I), (II):$Add and subtract terms

$∣∣ x=210 ∣∣ x=2-4 $

MoveNegNumToFrac

$(II):$ Put minus sign in front of fraction

$∣∣ x=210 ∣∣ x=-24 $

$(I), (II):$Calculate quotient

$x_{1}=5x_{2}=-2 $

Substitute | Simplify | |
---|---|---|

$x_{1}=5$ | $5+-5+14 =?2$ | $8 =2×$ |

$x_{2}=-2$ | $-2+-(-2)+14 =?2$ | $2=2✓$ |

Note that solving the equation by using inverse operations and Properties of Equality produced two solutions, $x=5$ and $x=-2.$ However, $x=5$ does not satisfy the original equation. Therefore, it is an extraneous solution. Conversely, since $x=-2$ satisfies the original equation, it is a solution. As a result, only Ali is correct and LaShay is incorrect.

Radical equations can be solved algebraically and graphically. Solving them algebraically sometimes produces extraneous solutions, whereas solving them graphically does not produce extraneous solutions. However, while algebraic solutions are generally exact, graphical solutions are often approximated. Consider an example radical equation.
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$x+5 =x+3 $

To solve this equation graphically, three steps must be followed.
1

Write Two Functions

Start by writing two functions. Each side of the equation represents a function.

$x+5 =x+3⇒y=x+5 y=x+3 $

The first is a radical function and the second is a linear function. 2

Graph Both Functions on the Same Coordinate Plane

To graph the functions, make a table of values for each. Remember that the radicand cannot be negative!

$y=x+5 $ | $y=x+3$ | |||
---|---|---|---|---|

$x$ | $x+5 $ | $y$ | $x+3$ | $y$ |

$-5$ | $-5+5 $ | $0$ | $-5+3$ | $-2$ |

$-4$ | $-4+5 $ | $1$ | $-4+3$ | $-1$ |

$-1$ | $-1+5 $ | $2$ | $-1+3$ | $2$ |

$4$ | $4+5 $ | $3$ | $4+3$ | $7$ |

Now the points for each function found in the table will be plotted on the same coordinate plane. Then the points of the linear function will be connected with a straight line, and the points of the radical function will be connected with a smooth curve.

3

Identify the Point of Intersection

The $x-$coordinate of the point of intersection of the functions gives the solution of the original equation.

The curve and the line intersect at $(-1,2).$ This means that the solution to the equation is $x=-1.$

After seeing LaShay's hard work, her teacher gives her extra credit homework so that she can have more practice with radical equations.

LaShay knows that if she solves this equation graphically, she will not have any extraneous solutions. Solve the radical equation graphically and find the value of $x.${"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.43056em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\">x<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["11"]}}

To solve a radical equation graphically, both sides of the equation need to be represented by functions.

$2x−6 =1+x−2 ⇓y=2x−6 andy=1+x−2 $

Now that there are two radical functions, they can be graphed by making a table of values. Since the radicands cannot be negative, the domain of each function will be found.
$Domain ofy=2x−6 2x−6≥0⇕x≥3 Domain ofy=1+x−2 x−2≥0⇕x≥2 $

Therefore, the table of values for the first function will be made for values of $x$ greater than or equal to $3.$ For the second function, the $x-$values must be greater than or equal to $2.$ $y=2x−6 $ | $y=1+x−2 $ | |||
---|---|---|---|---|

$x$ | $2x−6 $ | $y$ | $1+x−2 $ | $y$ |

$2$ | - | - | $1+2−2 $ | $1$ |

$3$ | $2(3)−6 $ | $0$ | $1+3−2 $ | $2$ |

$5$ | $2(5)−6 $ | $2$ | $1+5−2 $ | $≈2.73$ |

$6$ | $2(6)−6 $ | $≈2.45$ | $1+6−2 $ | $3$ |

$11$ | $2(11)−6 $ | $4$ | $1+11−2 $ | $4$ |

Now the points found in both functions can be plotted in the same coordinate plane. Then they can be connected with smooth curves.

Finally, the $x-$coordinate of the point of intersection of the curves gives the solution of the original equation.

Because the point of intersection is at $(11,4),$ the solution to the equation is $x=11.$

A radical inequality is an inequality that has an independent variable in a radicand or an independent variable with a rational exponent.

Variable in a Radicand | Variable With a Rational Exponent |
---|---|

$x−3 <x+3$ | $x_{21}>x−7$ |

$3x >2x+1$ | $1−x≤(x+1)_{31}$ |

$3≥44x+1 $ | $(3x)_{41}<9x−4$ |

This type of inequality can be solved algebraically.

Radical inequalities can be solved in a similar way to radical equations. The difference is that here, if the index is even, the $x-$values that make the radicand non-negative must be identified first. Consider an example radical inequality.
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$3x+9 −2≤7 $

To solve this inequality, five steps must be followed.
1

Identify the Values that Make the Radicand Non-negative

2

Isolate the Radical

Next, isolate the radical by using inverse operations.

$3x+9 −2≤7⇕3x+9 ≤9 $

3

Eliminate the Radical Symbol

4

Solve the Inequality

When the radical has been eliminated, the resulting inequality can be solved.
The values of $x$ that satisfy the given inequality are less than or equal to $24.$ Recall that at the beginning it was found that $x$ must be greater than or equal to $-3.$ Therefore, the solution set of the inequality can be written as a compound inequality.

$-3≤x≤24 $

5

Test Values to Check the Solution

Finally, some $x-$values can be tested to confirm the solution. In this case, three test values will be used. The first value will be less than $-3,$ the second greater than or equal to $-3$ and less than or equal to $24,$ and the third value will be greater than $24.$ In this case, $-4,$ $0,$ and $25$ will be tested.

$3x+9 −2≤7$ | ||
---|---|---|

$x$ | Substitute | Simplify |

$-4$ | $3(-4)+9 −2≤? 7$ | $-3 ≰7×$ |

$0$ | $3(0)+9 −2≤? 7$ | $1≤7✓$ |