Question

How do you solve `2b = 3+ \sqrt { 2b + 3}`?

Answer 1

`b = 3`

Explanation:

`2b = 3 + sqrt(2b+3)`

First, subtract `color(blue)3` from both sides of the equation:
`2b quadcolor(blue)(-quad3) = 3 + sqrt(2b+3) quadcolor(blue)(-quad3)`

`2b - 3 = sqrt(2b+3)`

Square both sides :
`(2b-3)^2 = sqrt(2b+3)^2`

To simplify the left hand side, we know that:

Following this image, it becomes:
`4b^2 - 12b + 9 = 2b+3`

Subtract `color(blue)3` from both sides :
`4b^2 - 12b +9 quadcolor(blue)(-quad3) = 2b + 3 quadcolor(blue)(-quad3)`

`4b^2 - 12b + 6 = 2b`

N ow subtract `color(blue)(2b)` from both sides :
`4b^2 - 12b + 6 quadcolor(blue)(-quad2b) = 2b quadcolor(blue)(-quad2b)`

`4b^2 - 14b + 6 = 0`

This is currently in the form `color(red)(a)x^2 + color(green)(b)x + color(blue)(c)`, where `color(red)(a = 4)`, `color(green)(b = -14)`, and `color(blue)(c = 6)`.

We have to factor this using two rules :
2 numbers must :

• Add up to `color(green)b`, or `color(green)(-14)`
• Multiply up to `color(red)a*color(blue)c`, or `color(red)(4)(color(blue)(6)) = 24`

Those two numbers are `-2` and `-12` .

So now:
`4b^2 - 2b - 12b + 6 = 0`

Factor by grouping :
`(4b^2 - 2b) + (-12b + 6) = 0`

`2b(2b - 1) -6(-2b - 1) = 0`

`(2b-6)(2b-1) = 0`

Since both expressions are multiplied to equal zero, that means we can set each expression equal to zero :

`2b - 6 = 0` and `2b - 1 = 0`

Simplify:
`2b = 6` `quadquadquad` and `quadquadquadquad2b = 1`

`b = 3` `quadquadquadquad` and `quadquadquadquadquadb = 1/2`

`--------------------`

We still need to check if they are really solutions by plugging them back into the original equation for `b` . First, let's check `3`:
`2b = 3 + sqrt(2b+3)`

`2(3) = 3 + sqrt(2(3)+3)`

`6 = 3 + sqrt(6+3)`

`3 = sqrt9`

`3 = 3`
Yes, this is a solution.

Now check `1/2`:
`2b = 3 + sqrt(2b+3)`

`2(1/2) = 3 + sqrt(2(1/2)+3)`

`1 = 3 + sqrt(1+3)`

`-2 = sqrt4`

`-2 != 2`
No, this is not a solution.

Therefore, `b = 3` .

Hope this helps!

Answer 2

`b=3`

Explanation:

Make the expression in the square roots on one side and other terms in the other side.

Then squaring on both sides

`(2b-3)^2=(sqrt(2b+3))^2`
`=>(2b-3)^2=2b+3`

Then expand the terms

`4b^2-12b+9=2b+3`

Take all terms of the equation to one side

`4b^2-12b-2b+9-3=0`
`=>4b^2-14b+6=0`

Then factorize the equation

`(2b-1)(b-3)=0`

After that solve

1. `2b-1=0=>b=1/2`

2. `b-3=0=>b=3`

So we got two value for `'b'` as `1/2and 3`

Check whether they are correct by substituting them in the given equation.

1. SUBSTITUTING " `color(red)(1/2` " in the equation

`2(1/2)=3+sqrt(2(1/2)+3`
`=>1=3+sqrt(4)`
`1=3+|2|`
`=>1=3+2`

`=>1=5` which is absurd

So `1/2` cannot be the answer

2.SUBSTITUTING "`color(red)(3)`" in the equation

`2(3)=3+sqrt(2(3)+3`
`=>6=3+sqrt9`
`6=6`

Hence `b=3` is the solution

Here is the link how to factor a equation
1.
How do you factor the expression `x^2-12x+27`?

2.How to factor any quadratic equation

`color(blue)(NOTE)`

Some people consider `sqrt(x^2)=+-x`, but it is wrong

In fact `sqrt(x^2)=|x|(` Absolute value of`'x'`)

Introduction to Absolute value function