Question

Which of the following are solutions to the equation below?

(2x+3)^2 = 10

Check all that apply.

Answer 1

Answer:

x = (√10  -3)/2  and (-√10  -3)/2

Step-by-step explanation:

(2x+3)^2 = 10

To solve the equation, take the square root of each side

sqrt((2x+3)^2) = ±√10

2x+3 = ±√10

Subtract 3 from each side

2x+3-3 = ±√10  -3

2x = ±√10  -3

Divide each side by 2

2x/2 = (±√10  -3)/2

x = (±√10  -3)/2

There are two solutions

x = (√10  -3)/2

and (-√10  -3)/2

       

Answer 2

Answer:

$\large {\textsf{A and D}}\ \implies \sf \sf \bold{x_1}=\dfrac{-\sqrt{10}-3}{2},\ \bold{x_2}=\dfrac{\sqrt{10}-3}{2}$

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.

$\sf \sqrt{(2x + 3)^2} = \sqrt{10}\\\\\Rightarrow 2x+3=\pm\sqrt{10}$

Step 2: Separate into possible cases.

$\sf x_1 \implies 2x+3=-\sqrt{10}\\\\x_2 \implies 2x+3=\sqrt{10}$

Step 3: Solve for x in both cases.

$\sf \bold{x_1} \implies 2x+3=-\sqrt{10}\ \ \textsf{[ Subtract 3 from both sides. ]}\\\\\Rightarrow 2x+3-3=-\sqrt{10}-3\\\\\Rightarrow 2x=-\sqrt{10}-3\ \ \textsf{[ Divide both sides by 2. ]}\\\\\Rightarrow \dfrac{2x}{2}=\dfrac{-\sqrt{10}-3}{2}\\\\\Rightarrow x_1=\dfrac{-\sqrt{10}-3}{2}$

$\sf \bold{x_2}\implies 2x+3=\sqrt{10}\ \ \textsf{[ Subtract 3 from both sides. ]}\\\\\Rightarrow 2x+3-3=\sqrt{10}-3\\\\\Rightarrow 2x=\sqrt{10}-3\ \ \textsf{[ Divide both sides by 2. ]}\\\\\Rightarrow \dfrac{2x}{2}=\dfrac{\sqrt{10}-3}{2}\\\\\Rightarrow x_2=\dfrac{\sqrt{10}-3}{2}$

Therefore, the solutions to this quadratic equation are: $\sf \bold{x_1}=\dfrac{-\sqrt{10}-3}{2},\ \bold{x_2}=\dfrac{\sqrt{10}-3}{2}$

Learn more about quadratic equations here:
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