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Equations with variable terms inside a radical are called radical equations. Like all equations, these can be solved graphically and algebraically. Sometimes, when solving algebraically, *extraneous solutions* — or solutions that do not satisfy the equation — arise.

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

*expand_more*

1

Write Two Functions

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

Show Solution *expand_more*

When solving an equation graphically it is necessary to have all variables on one side. Therefore, we'll first rearrange the equation.
The expression on the left-hand side can now be seen as the function,
We'll graph the function.
Thus, x=3 is a solution to the equation.

Next, we can identify the points that have the y-coordinate -1, then find the corresponding x-coordinates.

Since the point (3,-1) is a point on f, x=3 is a solution to the equation. We can verify this by testing it in the equation.$32x+2 =x−1$

Substitute

x=3

$32⋅3+2 =?3−1$

Multiply

Multiply

$36+2 =?3−1$

AddSubTerms

Add and subtract terms

$38 =?2$

CalcRoot

Calculate root

2=2

Radical equations can be solved algebraically using inverse operations. Specifically, to undo the radical, both sides of the equation can be raised to the same power as the index of the radical. For example,

Consider the following equation as an example.

1

Isolate the radical on one side

When solving a radical equation, it is necessary to isolate the radical on one side before raising the equation to an exponent. Using inverse operations, the equation becomes

2

Eliminate the radical

Now, the radical can be undone or eliminated by raising it to the same power as the index of the radical. Here, the radical is a **square** root, so the index is 2. Thus, both sides of the equation are raised to the power of 2.

3

Solve the equation

When the radical has been eliminated the resulting equation can be solved.
Now, the equation can be solved for the variable. Notice here that the highest exponent on x is 2. That means this is a quadratic equation, and must be solved as such. It'll be set equal to 0 before using the Quadratic Formula.
The x-values for the solutions to the equation are x=5 and x=1.

(2−x)2=2x−1

ExpandNegPerfectSquare

(a−b)2=a2−2ab+b2

22−2⋅2x+x2=2x−1

CalcPowProd

Calculate power and product

4−4x+x2=2x−1

4−4x+x2=2x−1

SubEqn

LHS−2x=RHS−2x

4−6x+x2=-1

AddEqn

LHS+1=RHS+1

5−6x+x2=0

RearrangeEqn

Rearrange equation

x2−6x+5=0

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⋅16±(-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 $

AddSubTerms

Add and subtract terms

$x=210 x=22 $

CalcQuot

Calculate quotient

$x=5x=1 $

4

Check for extraneous solutions

The solutions found in Step 3 might be extraneous solutions. Thus, they must be verified in the original equation. First the solution x=5 is tested.
Since we arrived at a contradiction, x=5 does not satisfy the radical equation. Thus, it is an extraneous solution. x=1 can be checked in the same way.

$2=x+2x−1 $

Substitute

x=5

$2=?5+2⋅5−1 $

Multiply

Multiply

$2=?5+10−1 $

SubTerms

Subtract terms

$2=?5+9 $

CalcRoot

Calculate root

$2=?5+3$

AddTerms

Add terms

2≠8

Since x=1 makes a true statement, it is a solution to the radical equation. With regard to discussing solutions, it can be said that $2=x+2x−1 $ has one solution and one extraneous solution.

Show Solution *expand_more*

Here, the Greek letters gamma, $γ,$ and beta, $β,$ are used to denote the variables. We can solve for $β$ by substituting $γ=1.25$ into the equation and using inverse operations to isolate $β.$
To eliminate the radical, both sides of the equation can be squared.
Lastly, we can determine if $β=0.6$ is an extraneous solution.
As $β=0.6$ made a true statement, it is a solution to the equation.

$γ=1−β_{2} 1 $

Substitute

$γ=1.25$

$1.25=1−β_{2} 1 $

MultEqn

$LHS⋅1−β_{2} =RHS⋅1−β_{2} $

$1.25⋅1−β_{2} =1$

DivEqn

$LHS/1.25=RHS/1.25$

$1−β_{2} =1.251 $

UseCalc

Use a calculator

$1−β_{2} =0.8$

$1−β_{2} =0.8$

RaiseEqn

$LHS_{2}=RHS_{2}$

$(1−β_{2} )_{2}=0.8_{2}$

PowSqrt

$(a )_{2}=a$

$1−β_{2}=0.8_{2}$

CalcPow

Calculate power

$1−β_{2}=0.64$

SubEqn

LHS−1=RHS−1

$-β_{2}=-0.36$

DivEqn

$LHS/(-1)=RHS/(-1)$

$β_{2}=0.36$

SqrtEqn

$LHS =RHS $

$β=±0.36 $

$β>0$

$β=0.36 $

CalcRoot

Calculate root

$β=0.6$

$1.25=1−β_{2} 1 $

Substitute

$β=0.6$

$1.25=?1−0.6_{2} 1 $

CalcPow

Calculate power

$1.25=?1−0.36 1 $

SubTerm

Subtract term

$1.25=?0.64 1 $

CalcRoot

Calculate root

$1.25=?0.81 $

UseCalc

Use a calculator

1.25=1.25

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