To find all real solutions for the equation [tex]\((2p + 3)^2 - 27 = 0\)[/tex], we will proceed step by step:
1. Start with the given equation:
[tex]\[(2p + 3)^2 - 27 = 0\][/tex]
2. Isolate the squared term:
To do this, add 27 to both sides:
[tex]\[(2p + 3)^2 = 27\][/tex]
3. Take the square root of both sides:
[tex]\[
2p + 3 = \pm \sqrt{27}
\][/tex]
Recall that [tex]\(\sqrt{27}\)[/tex] can be simplified to [tex]\(\sqrt{3^3} = 3\sqrt{3}\)[/tex]. Thus, the equation becomes:
[tex]\[
2p + 3 = \pm 3\sqrt{3}
\][/tex]
4. Solve for [tex]\(p\)[/tex]:
Divide the equation into two separate cases:
- Case 1: [tex]\(2p + 3 = 3\sqrt{3}\)[/tex]
[tex]\[
2p = 3\sqrt{3} - 3
\][/tex]
[tex]\[
p = \frac{3\sqrt{3} - 3}{2}
\][/tex]
Simplify the fraction:
[tex]\[
p = \frac{3(\sqrt{3} - 1)}{2}
\][/tex]
- Case 2: [tex]\(2p + 3 = -3\sqrt{3}\)[/tex]
[tex]\[
2p = -3\sqrt{3} - 3
\][/tex]
[tex]\[
p = \frac{-3\sqrt{3} - 3}{2}
\][/tex]
Simplify the fraction:
[tex]\[
p = \frac{-3(\sqrt{3} + 1)}{2}
\][/tex]
5. Write down the final solutions:
[tex]\[
p = \frac{3(\sqrt{3} - 1)}{2} \quad \text{and} \quad p = \frac{-3(\sqrt{3} + 1)}{2}
\][/tex]
In summary, the solutions are:
[tex]\[
p = -\frac{3}{2} + \frac{3\sqrt{3}}{2} \quad \text{and} \quad p = -\frac{3\sqrt{3}}{2} - \frac{3}{2}
\][/tex]
These are the two real solutions to the given equation.
