Understanding Radical Equations and Techniques for Solving Radical Inequalities

Variable in a Radicand	Variable With a Rational Exponent
$x - 2 = x + 2$	$x^{\frac{1}{2}} = 3 x - 5$
$3 2 x = x + 1$	$1 - 3 x = (x - 1)^{\frac{1}{3}}$
$3 = 4 2 x + 1$	$(2 x)^{\frac{1}{4}} = 4 x - 9$

Variable in a Radicand

Variable With a Rational Exponent

x - 2 = x + 2

x^{\frac{1}{2}} = 3 x - 5

3 2 x = x + 1

1 - 3 x = (x - 1)^{\frac{1}{3}}

3 = 4 2 x + 1

(2 x)^{\frac{1}{4}} = 4 x - 9

	Substitute	Simplify
$x = 1$	$2 = ? 1 + 2 (1) - 1$	$2 = 2$ $✓$
$x = 5$	$2 = ? 5 + 2 (5) - 1$	$2 \neq = 8$ $\times$

Substitute

Simplify

x = 1

2 = ? 1 + 2 (1) - 1

2 = 2

✓

x = 5

2 = ? 5 + 2 (5) - 1

2 \neq = 8

\times

	Substitute	Simplify
$x_{1} = 5$	$5 + - 5 + 14 = ? 2$	$8 \neq = 2 \times$
$x_{2} = - 2$	$- 2 + - (- 2) + 14 = ? 2$	$2 = 2 ✓$

Substitute

Simplify

x_{1} = 5

5 + - 5 + 14 = ? 2

8 \neq = 2 \times

x_{2} = - 2

- 2 + - (- 2) + 14 = ? 2

2 = 2 ✓

	$y = x + 5$	$y = x + 3$
$x$	$x + 5$	$y$	$x + 3$	$y$
$- 5$	$- 5 + 5$	$0$	$- 5 + 3$	$- 2$
$- 4$	$- 4 + 5$	$1$	$- 4 + 3$	$- 1$
$- 1$	$- 1 + 5$	$2$	$- 1 + 3$	$2$
$4$	$4 + 5$	$3$	$4 + 3$	$7$

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

	$y = 2 x - 6$	$y = 1 + x - 2$
$x$	$2 x - 6$	$y$	$1 + x - 2$	$y$
$2$	-	-	$1 + 2 - 2$	$1$
$3$	$2 (3) - 6$	$0$	$1 + 3 - 2$	$2$
$5$	$2 (5) - 6$	$2$	$1 + 5 - 2$	$\approx 2.73$
$6$	$2 (6) - 6$	$\approx 2.45$	$1 + 6 - 2$	$3$
$11$	$2 (11) - 6$	$4$	$1 + 11 - 2$	$4$

y = 2 x - 6

y = 1 + x - 2

x

2 x - 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

\approx 2.73

6

2 (6) - 6

\approx 2.45

1 + 6 - 2

3

11

2 (11) - 6

4

1 + 11 - 2

4

Variable in a Radicand	Variable With a Rational Exponent
$x - 3 < x + 3$	$x^{\frac{1}{2}} > x - 7$
$3 x > 2 x + 1$	$1 - x \leq (x + 1)^{\frac{1}{3}}$
$3 \geq 4 4 x + 1$	$(3 x)^{\frac{1}{4}} < 9 x - 4$

Variable in a Radicand

Variable With a Rational Exponent

x - 3 < x + 3

x^{\frac{1}{2}} > x - 7

3 x > 2 x + 1

1 - x \leq (x + 1)^{\frac{1}{3}}

3 \geq 4 4 x + 1

(3 x)^{\frac{1}{4}} < 9 x - 4

$3 x + 9 - 2 \leq 7$
$x$	Substitute	Simplify
$- 4$	$3 (- 4) + 9 - 2 \leq ? 7$	$- 3 ≰ 7 \times$
$0$	$3 (0) + 9 - 2 \leq ? 7$	$1 \leq 7 ✓$
$25$	$3 (25) + 9 - 2 \leq ? 7$	$7.2 ≰ 7 \times$

3 x + 9 - 2 \leq 7

x

Substitute

Simplify

- 4

3 (- 4) + 9 - 2 \leq ? 7

- 3 ≰ 7 \times

0

3 (0) + 9 - 2 \leq ? 7

1 \leq 7 ✓

25

3 (25) + 9 - 2 \leq ? 7

7.2 ≰ 7 \times

$2 x - 4 + 4 < 8$
$x$	Substitute	Simplify
$0$	$2 (0) - 4 + 4 < 8$	$- 4 ≮ 4 \times$
$4$	$2 (4) - 4 + 4 < 8$	$6 < 8 ✓$
$20$	$2 (20) - 4 + 4 < 8$	$10 ≮ 8 \times$

2 x - 4 + 4 < 8

x

Substitute

Simplify

0

2 (0) - 4 + 4 < 8

- 4 ≮ 4 \times

4

2 (4) - 4 + 4 < 8

6 < 8 ✓

20

2 (20) - 4 + 4 < 8

10 ≮ 8 \times

$10 t + 2 t = 10 2 t$
$t$	Substitute	Simplify
$0$	$10 0 + 2 (0) = ? 10 2 (0)$	$0 = 0 ✓$
$2$	$10 2 + 2 (2) = ? 10 2 (2)$	$20 = 20 ✓$

10 t + 2 t = 10 2 t

t

Substitute

Simplify

0

10 0 + 2 (0) = ? 10 2 (0)

0 = 0 ✓

2

10 2 + 2 (2) = ? 10 2 (2)

20 = 20 ✓

Solve the radical equation.

x + 9 - x + 1 = 2

Solving a radical equation usually involves three main steps.

Isolate the radical on one side of the equation.
Raise each side of the equation to a power that is equal to radical's index to eliminate the radical.
Solve the resulting equation. Remember to check the results!

Now we can analyze the given radical equation. sqrt(x+9)-sqrt(x+1)=2 First, let's isolate the radical sqrt(x+1) on one side of the equation. sqrt(x+9)-sqrt(x+1)=2 ⇕ sqrt(x+9)=sqrt(x+1)+2 We get an isolated radical with index equal to 2. Then, we will raise each side of the equation to the power of 2.

We got another radical equation. Let's now isolate the radical sqrt(x+1) on one side of this new equation.

Now, just like before, we will raise each side of the equation to the power of 2.

Next, we will check the solution by substituting 0 for x in the original equation. If the substitution produces a true statement, we know that our answer is correct. Otherwise, x=0 is an extraneous solution.

Because our substitution produced a true statement, we know that our answer correct.

What is the solution to the radical equation?

x + 3 + 3 = x

To solve equations with a variable expression inside a radical, we first want to make sure the radical is isolated. Then we can raise both sides of the equation to a power equal to the index of the radical. Let's try to solve our equation using this method!

Finding the Solutions

Let's take another look at the given equality. sqrt(x+3)+3=x We will first isolate the radical with the variable term inside and then raise each side of the equation to the power of 2.

We now have a quadratic equation and we want to find its solutions. To do so, let's identify the values of a, b, and c. x^2-7x+6=0 ⇓ 1x^2+( -7)x+ 6=0 We can see that a= 1, b= - 7, and c= 6. Let's substitute these values into the Quadratic Formula.

Using the Quadratic Formula, we found that the solutions are x_1= 6 and x_2= 1. Let's check them to see if we have any extraneous solutions.

Checking the Solutions

We will check x_1=6 and x_2=1 one at a time. Let's start by substituting 1 for x into the original equation.

We got a false statement, so x=1 is an extraneous solution. Let's now substitute 6 for x and evaluate the equation.

In this case we got a true statement. Therefore, x=6 is the only solution to the original equation.

What is the solution to the following radical inequality?

2 - b + 8 \leq - b

Write the answer as a compound inequality.

Before solving the radical inequality, recall that the radicand of a square root must be greater than or equal to 0. sqrt(2)-sqrt(b+8)≤ -sqrt(b) Let's start by identifying the values of b for which the inequality is defined.

Radical	Radicand Greater Than or Equal to Zero	Condition for the Variable
sqrt(b+8)	b+8≥ 0	b≥ - 8
sqrt(b)	b≥ 0	b≥ 0

After solving the given inequality, we will combine the solution with the above intervals to obtain the solution set. Let's start by isolating one of the radicals.

Our next step is to raise both sides of the above inequality to the power of 2. To do this, we have to be sure both sides are non-negative, because we may get incorrect solutions otherwise. In this case, both sides are be non-negative because square roots are always non-negative. This means that we can proceed with squaring both sides of the inequality.

Now we can combine the three intervals.

The intersection of the three solutions, which is the solution set of the given inequality, is 0≤ b≤ 4.5.

LaShay is working on an equation that involves a rational exponent. She knows that she made a mistake when solving the equation but she is not certain where.

Correct the error and find the solution to the equation.

LaShay's error lies in the first step. To eliminate the power, both sides of the equation must be raised to the reciprocal of the variable's exponent. However, whatever operation is performed, must be applied to the entirety of the left- and right-hand sides. 27(x^()32)^()23=216^()23 & * ( 27x^()32)^()23=216^()23 & ✓ Now we can solve the equation correctly.

In a hurricane, the average sustained wind velocity

v,

in meters per second, can be modeled by the following equation.

v (p) = 6.3 1013 - p

In this equation,

p

is the air pressure, in millibars, at the center of the hurricane.

Estimate the air pressure at the center of the hurricane when the average sustained wind velocity is

45.4

meters per second. Round the answer to the nearest integer.

Given that the average sustained velocity is 45.5 meters per second, we will begin by substituting 45.4 for v(p) into the equation. Then we can solve it for p.

The air pressure at the center of the hurricane is about 961 millibars.

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