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Lesson Example Discussion Quiz: Class Homework

Step-3

Title: Equation with two radicals

Grade: 8-b Lesson: S2-L4

Explanation: Hello Students, time to practice and review the steps for the problem.

Lesson Steps

Discussion: Step1 Step2 Step3 Step4 Step5

Step	Type	Explanation	Answer
1	Problem	Solve the following radical equation \$\sqrt(3( 2s + 7) - 15) = \sqrt (5s + 2 + 4)\$.
2	Step	The given equation	\$\sqrt(3( 2s + 7) - 15) = \sqrt (5s + 2 + 4)\$
3	Hint	Square both sides of the equation to remove the square roots and simplify it:	\$(\sqrt(3( 2s + 7) - 15))^2 = (\sqrt (5s + 6))^2\$ \$3(2s +7) - 15 = 5s +6\$
4	Step	Expand and simplify the left-hand side:	\$ 3(2s + 7) - 15 = 6s + 21 -15\$ \$3(2s + 7) - 15 = 6s + 6 \$
5	Step	So, the equation simplifies to:	6s + 6 = 5s + 6
6	Step	Isolate s by moving all terms involving s to one side and the constants to the other:	6s - 5s = 6 - 6 s = 0
7	Step	So, the solution to the equation is: s = 0.
8	Choice.A	Substituting s = 1 into the original equations reveals that it does not satisfy them so it is wrong	s = 1
9	Choice.B	If s = 3 is substituted into the original equations, it does not satisfy them so it is wrong	s = 3
10	Choice.C	Substituting s = 2 into the original equations does not satisfy them, indicating it’s not a solution	s = 2
11	Choice.D	Substitute s = 0 into the original equations; it satisfies them, proving it as a valid solution	s = 0
12	Answer	Option	D
13	Sumup	Can you summarize what you’ve understood in the above steps?

Step

Type

Explanation

Answer

Problem

Solve the following radical equation \$\sqrt(3( 2s + 7) - 15) = \sqrt (5s + 2 + 4)\$.

Step

The given equation

\$\sqrt(3( 2s + 7) - 15) = \sqrt (5s + 2 + 4)\$

Hint

Square both sides of the equation to remove the square roots and simplify it:

\$(\sqrt(3( 2s + 7) - 15))^2 = (\sqrt (5s + 6))^2\$

\$3(2s +7) - 15 = 5s +6\$

Step

Expand and simplify the left-hand side:

\$ 3(2s + 7) - 15 = 6s + 21 -15\$

\$3(2s + 7) - 15 = 6s + 6 \$

Step

So, the equation simplifies to:

6s + 6 = 5s + 6

Step

Isolate s by moving all terms involving s to one side and the constants to the other:

6s - 5s = 6 - 6

s = 0

Step

So, the solution to the equation is: s = 0.

Choice.A

Substituting s = 1 into the original equations reveals that it does not satisfy them so it is wrong

s = 1

Choice.B

If s = 3 is substituted into the original equations, it does not satisfy them so it is wrong

s = 3

Choice.C

Substituting s = 2 into the original equations does not satisfy them, indicating it’s not a solution

s = 2

Choice.D

Substitute s = 0 into the original equations; it satisfies them, proving it as a valid solution

s = 0

Answer

Option

Sumup

Can you summarize what you’ve understood in the above steps?

Discussion: Step1 Step2 Step3 Step4 Step5