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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.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 | x21=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.
Substitute | Simplify | |
---|---|---|
x=1 | 2=?1+2(1)−1 | 2=2 ✓ |
x=5 | 2=?5+2(5)−1 | 2=8 × |
(a−b)2=a2−2ab+b2
Calculate power and product
LHS−2x=RHS−2x
LHS+1=RHS+1
Commutative Property of Addition
Use the Quadratic Formula: a=1,b=-6,c=5
-(-a)=a
Calculate power and product
Subtract term
Calculate root
State solutions
(I), (II): Add and subtract terms
(I), (II): Calculate quotient
x=1
Identity Property of Multiplication
Subtract term
Calculate root
Add terms
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.The perimeter of a rectangle is P=2ℓ+2w, where ℓ is the length and w is the width of the rectangle.
Multiply
LHS−16=RHS−16
LHS/2=RHS/2
Rearrange equation
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.Substitute the given values into the equation and solve for h.
LHS−1=RHS−1
LHS⋅4=RHS⋅4
LHS2=RHS2
(a)2=a
LHS+h=RHS+h
LHS−64=RHS−64
h=36
Subtract term
Calculate root
b1⋅a=ba
Calculate quotient
\AddTerm
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.Start by isolating the radical expression.
LHS2=RHS2
(a)2=a
(a−b)2=a2−2ab+b2
LHS+x=RHS+x
LHS−14=RHS−14
Commutative Property of Addition
Rearrange equation
Use the Quadratic Formula: a=1,b=-3,c=-10
-(-a)=a
Calculate power and product
Add terms
Calculate root
State solutions
(I), (II): Add and subtract terms
(II): Put minus sign in front of fraction
(I), (II): Calculate quotient
Substitute | Simplify | |
---|---|---|
x1=5 | 5+-5+14=?2 | 8=2 × |
x2=-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.
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.
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.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 | x21>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.
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 ✓ |
25 | 3(25)+9−2 |