Asked by local.trin
Show how many solutions are in the solution set for the equation 3x2=21.(1 point) Responses zero zero two two one one infinitely many
Answers
Answered by
local.trin
Show how many solutions are in the solution set for 2(x−5)2+2=20.(1 point) Responses one one infinitely many infinitely many two two zero
Answered by
local.trin
Solve x2=16/25 by inspection. There are two real solutions. Enter the lesser number first. Leave the answers in simplest fraction form
Answered by
GPT-4o mini
To solve the equation \( 3x^2 = 21 \), we first divide both sides by 3:
\[
x^2 = \frac{21}{3} = 7
\]
Next, we take the square root of both sides:
\[
x = \pm \sqrt{7}
\]
This means there are two solutions:
\[
x = \sqrt{7} \quad \text{and} \quad x = -\sqrt{7}
\]
Therefore, the number of solutions in the solution set for the equation \( 3x^2 = 21 \) is **two**.
\[
x^2 = \frac{21}{3} = 7
\]
Next, we take the square root of both sides:
\[
x = \pm \sqrt{7}
\]
This means there are two solutions:
\[
x = \sqrt{7} \quad \text{and} \quad x = -\sqrt{7}
\]
Therefore, the number of solutions in the solution set for the equation \( 3x^2 = 21 \) is **two**.
Answered by
GPT-4o mini
To solve the equation \( 2(x-5)^2 + 2 = 20 \), we first isolate the term with the squared expression.
1. Subtract 2 from both sides:
\[
2(x-5)^2 = 20 - 2
\]
\[
2(x-5)^2 = 18
\]
2. Divide both sides by 2:
\[
(x-5)^2 = \frac{18}{2} = 9
\]
3. Next, take the square root of both sides:
\[
x - 5 = \pm 3
\]
This leads to two equations:
- \( x - 5 = 3 \)
- \( x - 5 = -3 \)
Now, we solve these equations:
1. For \( x - 5 = 3 \):
\[
x = 3 + 5 = 8
\]
2. For \( x - 5 = -3 \):
\[
x = -3 + 5 = 2
\]
Thus, the solutions are \( x = 8 \) and \( x = 2 \).
Therefore, the number of solutions in the solution set for the equation \( 2(x-5)^2 + 2 = 20 \) is **two**.
1. Subtract 2 from both sides:
\[
2(x-5)^2 = 20 - 2
\]
\[
2(x-5)^2 = 18
\]
2. Divide both sides by 2:
\[
(x-5)^2 = \frac{18}{2} = 9
\]
3. Next, take the square root of both sides:
\[
x - 5 = \pm 3
\]
This leads to two equations:
- \( x - 5 = 3 \)
- \( x - 5 = -3 \)
Now, we solve these equations:
1. For \( x - 5 = 3 \):
\[
x = 3 + 5 = 8
\]
2. For \( x - 5 = -3 \):
\[
x = -3 + 5 = 2
\]
Thus, the solutions are \( x = 8 \) and \( x = 2 \).
Therefore, the number of solutions in the solution set for the equation \( 2(x-5)^2 + 2 = 20 \) is **two**.
Answered by
GPT-4o mini
To solve the equation \( x^2 = \frac{16}{25} \) by inspection, we recognize that we can take the square root of both sides.
Taking the square root, we get:
\[
x = \pm \sqrt{\frac{16}{25}}
\]
Calculating the square root of the fraction:
\[
x = \pm \frac{\sqrt{16}}{\sqrt{25}} = \pm \frac{4}{5}
\]
Thus, the two real solutions are \( \frac{4}{5} \) and \( -\frac{4}{5} \).
Arranging these solutions in order, with the lesser number first, we have:
\[
-\frac{4}{5}, \quad \frac{4}{5}
\]
So the final answers are:
\(-\frac{4}{5}, \frac{4}{5}\)
Taking the square root, we get:
\[
x = \pm \sqrt{\frac{16}{25}}
\]
Calculating the square root of the fraction:
\[
x = \pm \frac{\sqrt{16}}{\sqrt{25}} = \pm \frac{4}{5}
\]
Thus, the two real solutions are \( \frac{4}{5} \) and \( -\frac{4}{5} \).
Arranging these solutions in order, with the lesser number first, we have:
\[
-\frac{4}{5}, \quad \frac{4}{5}
\]
So the final answers are:
\(-\frac{4}{5}, \frac{4}{5}\)
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