To solve the equation [tex]\((2x + 3)^2 = 10\)[/tex], we need to determine all possible values of [tex]\(x\)[/tex].
First, let's take the square root of both sides of the equation:
[tex]\[ (2x + 3)^2 = 10 \][/tex]
[tex]\[ 2x + 3 = \pm \sqrt{10} \][/tex]
We now have two cases to consider:
### Case 1: [tex]\( 2x + 3 = \sqrt{10} \)[/tex]
Solving for [tex]\(x\)[/tex]:
[tex]\[ 2x + 3 = \sqrt{10} \][/tex]
[tex]\[ 2x = \sqrt{10} - 3 \][/tex]
[tex]\[ x = \frac{\sqrt{10} - 3}{2} \][/tex]
### Case 2: [tex]\( 2x + 3 = -\sqrt{10} \)[/tex]
Solving for [tex]\(x\)[/tex]:
[tex]\[ 2x + 3 = -\sqrt{10} \][/tex]
[tex]\[ 2x = -\sqrt{10} - 3 \][/tex]
[tex]\[ x = \frac{-\sqrt{10} - 3}{2} \][/tex]
Therefore, the solutions to the equation [tex]\((2x + 3)^2 = 10\)[/tex] are:
1. [tex]\( x = \frac{\sqrt{10} - 3}{2} \)[/tex]
2. [tex]\( x = \frac{-\sqrt{10} - 3}{2} \)[/tex]
Now, let's check the given choices:
A. [tex]\( x = \sqrt{10} + \frac{3}{2} \)[/tex]
This does not match either solution.
B. [tex]\( x = \frac{-\sqrt{10} - 3}{2} \)[/tex]
This matches the second solution.
C. [tex]\( x = -\sqrt{10} + \frac{3}{2} \)[/tex]
This does not match either solution.
D. [tex]\( x = \frac{\sqrt{7}}{2} \)[/tex]
This value does not solve the original equation.
E. [tex]\( x = \frac{\sqrt{10} - 3}{2} \)[/tex]
This matches the first solution.
F. [tex]\( x = -\frac{\sqrt{7}}{2} \)[/tex]
This value does not solve the original equation.
Therefore, the correct solutions from the given choices are:
- B. [tex]\( x = \frac{-\sqrt{10} - 3}{2} \)[/tex]
- E. [tex]\( x = \frac{\sqrt{10} - 3}{2} \)[/tex]
