Answer:
[tex]\large {\textsf{A and D}}\ \implies \sf \sf \bold{x_1}=\dfrac{-\sqrt{10}-3}{2},\ \bold{x_2}=\dfrac{\sqrt{10}-3}{2}[/tex]
Step-by-step explanation:
Given: (2x + 3)² = 10
In order to find the solutions to the given equation, we can take the (square) roots of the equation to find the zeros, which are also known as the x-intercepts. This is where the zeros intersect the x-axis.
Note: when taking the square roots of a quadratic equation, remember to use both the positive and negative roots.
Step 1: Square both sides of the equation.
[tex]\sf \sqrt{(2x + 3)^2} = \sqrt{10}\\\\\Rightarrow 2x+3=\pm\sqrt{10}[/tex]
Step 2: Separate into possible cases.
[tex]\sf x_1 \implies 2x+3=-\sqrt{10}\\\\x_2 \implies 2x+3=\sqrt{10}[/tex]
Step 3: Solve for x in both cases.
[tex]\sf \bold{x_1} \implies 2x+3=-\sqrt{10}\ \ \textsf{[ Subtract 3 from both sides. ]}\\\\\Rightarrow 2x+3-3=-\sqrt{10}-3\\\\\Rightarrow 2x=-\sqrt{10}-3\ \ \textsf{[ Divide both sides by 2. ]}\\\\\Rightarrow \dfrac{2x}{2}=\dfrac{-\sqrt{10}-3}{2}\\\\\Rightarrow x_1=\dfrac{-\sqrt{10}-3}{2}\\\\[/tex]
[tex]\sf \bold{x_2}\implies 2x+3=\sqrt{10}\ \ \textsf{[ Subtract 3 from both sides. ]}\\\\\Rightarrow 2x+3-3=\sqrt{10}-3\\\\\Rightarrow 2x=\sqrt{10}-3\ \ \textsf{[ Divide both sides by 2. ]}\\\\\Rightarrow \dfrac{2x}{2}=\dfrac{\sqrt{10}-3}{2}\\\\\Rightarrow x_2=\dfrac{\sqrt{10}-3}{2}[/tex]
Therefore, the solutions to this quadratic equation are: [tex]\sf \bold{x_1}=\dfrac{-\sqrt{10}-3}{2},\ \bold{x_2}=\dfrac{\sqrt{10}-3}{2}[/tex]
Learn more about quadratic equations here:
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