Question

What values of b satisfy 3(2b + 3)^2 = 36?

Answer 1

Answer:
either b = -1.5 + √3
or b = -1.5 - √3

Explanation:
To solve this problem, we will simplify the expression on the left-hand side and solve for "b" as follows:
The given expression is:
3(2b+3)² = 36

1- Divide both sides of the equation by 3. This will give:
(2b+3)² = 12

2- Expand the bracket as follows:
(2b+3)² = 12
(2b)² + 2(2b)(3) + (3)² = 12
4b² + 12b + 9 = 12

3- Put the equation is standard form (ax² + bx + c = 0):
4b² + 12b + 9 = 12
4b² + 12b + 9 - 12 = 0
4b² + 12b - 3 = 0

4- Factorize the equation to get the values of "b":
4b² + 12b - 3 = 0
By comparing the given equation with the standard form, we will find that:
a = 4
b = 12 
c = -3
Use the quadratic formula shown in the attached image, substitute with the values of a, b and c and solve for "b"
This will give us:
either b = -1.5 + √3
or b = -1.5 - √3

Hope this helps :)

Answer 2

ANSWER

$b = - \sqrt{3} - \frac{3}{2} \: or \: \sqrt{3} - \frac{3}{2}$

EXPLANATION

The given equation is
$3 {(2b + 3)}^{2} = 36$

We can quickly solve this using the square root method.

We first of all divide both sides of the equation by 3 to obtain,

$\frac{3{(2b + 3)}^{2}}{3} = \frac{36}{3}$

This implies that,

${(2b + 3)}^{2} = 12$

We now take the square root of both sides to obtain,

$(2b + 3)= \pm \sqrt{12}$

This simplifies to,

$2b + 3 = \pm \: 2 \sqrt{3}$

We add the additive inverse of 3, which is -3 to both sides of the equation to obtain,

$2b = - 3\pm \: 2 \sqrt{3}$

We now divide through by 2 to obtain,

$b = - \frac{3}{2} \pm \: \sqrt{3}$

We split the plus or minus sign to obtain,

$b = - \frac{3}{2} + \: \sqrt{3}$

or

$b = - \frac{3}{2} - \: \sqrt{3}$