Question

1. What is the solution set of (3x - 1)2 = 5?

2. What is the solution set of x 2 + 5x + 1 = 0?

3. Click an item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the correct position in the answer box. Release your mouse button when the item is place. If you change your mind, drag the item to the trashcan. Click the trashcan to clear all your answers.
Solve this quadratic equation.

x 2 + 5x + 3 = 0

Answer 1

Question 1:

Consider the equation $(3x-1)^2 = 5$

Taking square root on both the sides of the equation, we get

$(3x-1) = \pm \sqrt{5}$

Consider $3x-1 = \sqrt 5$

adding '1' to both sides, we get

$3x = \sqrt 5 + 1$

Dividing by '3', we get

$x = \frac{\sqrt5 + 1}{3}$

Consider $3x-1 = - \sqrt 5$

adding '1' to both sides, we get

$3x = - \sqrt 5 + 1$

Dividing by '3', we get

$x = \frac{- \sqrt5 + 1}{3}$

So, the solution set for this equation is $x = \frac{\sqrt5 + 1}{3}$ and $x = \frac{- \sqrt5 + 1}{3}$.

Question 2:

Consider the equation $x^2+5x+1 = 0$

We will use quadratic formula, we get

$x = \frac{-5\pm \sqrt{5^2-4}}{2}$

$x = \frac{-5\pm \sqrt{21}}{2}$

So, the solution set for this equation is $x = \frac{-5+ \sqrt{21}}{2}, x = \frac{-5- \sqrt{21}}{2}$.

Question 3:

Consider the equation $x^2+5x+3=0$

By using the quadratic formula, we get

$x = \frac{-5\pm \sqrt{5^2-12}}{2}$

$x = \frac{-5\pm \sqrt{13}}{2}$

so, the solution set for this equation is $x = \frac{-5+ \sqrt{13}}{2}, x = \frac{-5- \sqrt{13}}{2}$.

Answer 2

Answer. B :

Step-by-step explanation:

Because I just took test