Question

Solve the equation. Check both solutions and only write the real solution(s). square root of 2x+3=x

Answer 1

Is this what you mean
\[\sqrt{2x + 3} = x \]

Answer 2

yes

Answer 3

We are going to solve it together.

First square both sides

that gives 2x + 3 = x^2

Answer 4

does that make sense

Answer 5

If it is, square both sides to get:\[x^{2} = 2x + 3 \rightarrow x^{2} - 2x - 3 = 0\]Remember that\[\Delta=b^{2} - 4ac\]In specific, we get:\[\Delta = 4 - (4*(-3)*1) = 16\]So the binomial has distinct real roots.

Answer 6

@denyekwe what do i do after that?

Answer 7

next you will want to subtract(2x + 3) from both sides of the equation.

Answer 8

I chose (2x + 3), because i wanted to arrive to

x^2 - 2x - 3 = 0

Answer 9

then do i solve after this ?

Answer 10

Yes, we are left with a quadratic equation.

Answer 11

do i end up with x^2-2x=0?

Answer 12

No

Answer 13

What did you do?

Answer 14

x^2-2x=0 then added 3 to both sides

Answer 15

If you add three to both sides we will have something like this.

x^2 - 2x - 3 + 3 = 0 + 3

Answer 16

Evaluating that will give us,

x^2 - 2x = 3

Answer 17

i'm sorry,
that is what i got

Answer 18

Thats fine. Everyone makes mistakes.

Answer 19

To solve

x^2 - 2x - 3 = 0

we have to treat it as a quadratic equation.

Answer 20

Are you familiar with quadratic equations

Answer 21

kinda

Answer 22

I would love to teach you somethings about quadratic equations but i have something else i am working on. This link " http://www.khanacademy.org/ " has lots of FREE videos on solving quadratic equations. I suggest you visit it and watch the lectures.

Answer 23

The list of videos is really long so i suggest that you do a find on the web page for the word "Algebra: Quadratics". This will take you to the section where videos for solving quadratic equations are located.

Answer 24

But to finish up the problem we had, i will use the quadratic formular.

Answer 25

(1)x^2 - (2)x - (3) = 0
a b c

The quadratic formula
\[x = (-b \pm \sqrt{b ^{2} - 4ac})\div2a\]
from our equation,

a = 1
b = -2
c = -3

\[x = (-(-2) \pm \sqrt{(-2) ^{2} - 4(1)(-3)})\div2(1)\]
\[x = ( 2\pm \sqrt{4 + 12})\div2\]
\[x = ( 2\pm \sqrt{16})\div2\]
\[x = ( 2\pm 4)\div2\]
this leaves us with

x = (2-4)/2 = -1
x = (2+4)/2 = 3

Answer 26

Going back to the original question.

(2x+3)^(0.5) = x

when we sub our x from above, we get

(2(-1)+3)^(0.5) = (-1)
(-2+3)^(0.5) = (-1)
(1)^(0.5) = (-1)
1 = -1 ### Since this statement is not true x = -1, does not belong to my solution.

(2(3)+3)^(0.5) = (3)
(6+3)^(0.5) = (3)
(9)^(0.5) = (3)
3 = 3 ### Since this statement is true x = 3, does belong to my solution.

Answer 27

There is a special trick to do this problem. You can show that
if a-b+c =0, then one root is -1 and the other is -c/a

On a related matter, if a+b+c=0 then one root is 1 and the other is c/a.

Both statement comes from the fact that the product of the two roots is c/a

Answer 28

You can practice problems of roots of polynomials on
http://www.saab.org/mathdrills/act.html

Answer 29

sqrt(2x+3)=x both sides squared

2x+3=x^2
x^2 -2x -3=0
(x-3)(x+1)=0
x-3=0
x=3

x+1=0
x=-1