Westonci.ca is the ultimate Q&A platform, offering detailed and reliable answers from a knowledgeable community. Get quick and reliable solutions to your questions from knowledgeable professionals on our comprehensive Q&A platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.

Identify the solutions and any extraneous roots.

1. [tex]\sqrt{15x - 10} = \sqrt{2x + 3}[/tex]


Sagot :

Certainly! Let's solve the equation step-by-step to identify the solutions and any extraneous solutions.

The equation given is:
[tex]\[ \sqrt{15x - 10} = \sqrt{2x + 3} \][/tex]

1. Square both sides to eliminate the square roots:
[tex]\[ (\sqrt{15x - 10})^2 = (\sqrt{2x + 3})^2 \][/tex]

2. This simplifies to:
[tex]\[ 15x - 10 = 2x + 3 \][/tex]

3. Isolate [tex]\( x \)[/tex] by moving all terms involving [tex]\( x \)[/tex] to one side and constants to the other:
[tex]\[ 15x - 2x = 3 + 10 \][/tex]

4. Combine like terms:
[tex]\[ 13x = 13 \][/tex]

5. Solve for [tex]\( x \)[/tex]:
[tex]\[ x = 1 \][/tex]

Now, we have a potential solution: [tex]\( x = 1 \)[/tex].

6. Verify the solution to see if it is extraneous:

Substitute [tex]\( x = 1 \)[/tex] back into the original equation:
[tex]\[ \sqrt{15(1) - 10} = \sqrt{2(1) + 3} \][/tex]
[tex]\[ \sqrt{15 - 10} = \sqrt{2 + 3} \][/tex]
[tex]\[ \sqrt{5} = \sqrt{5} \][/tex]

Since both sides of the equation are equal when [tex]\( x = 1 \)[/tex], this is a valid solution.

7. Conclusion:
The solution to the equation [tex]\( \sqrt{15x - 10} = \sqrt{2x + 3} \)[/tex] is:
[tex]\[ x = 1 \][/tex]

There are no extraneous solutions in this case.

Thus, the result is:
- Real Solutions: [tex]\[ x = 1 \][/tex]
- Extraneous Solutions: None