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Which of the following are solutions to the equation below? Check all that apply.

[tex]\((2x + 3)^2 = 10\)[/tex]

A. [tex]\(x = \sqrt{10} + \frac{3}{2}\)[/tex]

B. [tex]\(x = \frac{-\sqrt{10} - 3}{2}\)[/tex]

C. [tex]\(x = -\sqrt{10} + \frac{3}{2}\)[/tex]

D. [tex]\(x = \frac{\sqrt{7}}{2}\)[/tex]

E. [tex]\(x = \frac{\sqrt{10} - 3}{2}\)[/tex]

F. [tex]\(x = -\frac{\sqrt{7}}{2}\)[/tex]


Sagot :

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]