Question

Solve Sq. Root 2x+3 = x + 1 for x.
Could someone help me with this? (Picture included)

Answer 1

\[\sqrt{2x+3} = x + 1\]

Answer 2

Do you know how to get rid of a square root?

What do you have to do to both sides?

Answer 3

@marissalovescats Yes, I squared both sides and got 2x+3 = x^2 + 2x + 1

Answer 4

@marissalovescats I see this will be a quadratic equation, right?

Answer 5

You are correct, we will bring everything from the left to the right using algebra, which will leave us with 0=

Answer 6

@marissalovescats this is where I got stuck, to get 0 on one side, I had to -2x from both sides, leaving me with 0 = x^2 - 2.

Answer 7

Yup you sure do have that, now we can take the long way and do the quadratic formula, or we can add 2 to the left and have:

2=x^2 right?

Answer 8

Then we would take the square root of each side to solve for x right?

Leaving us with what answer

Answer 9

@marissalovescats Yes, the quadratic formula!! Thank you, that completely slipped my mind. I was really focused on factoring them out and was stuck when I didn't have an x in the quadratic formula!

Answer 10

If you do the quadratic formula, just make sure you use 0 as your b, because technically we have an equation of:

x^2+0x-2. Sometimes it helps to write it that way.

Answer 11

@marissalovescats Yes, so a = 1 b = 0 and c = -2. Thank you!

Answer 12

But we don't have to do the long quadratic formula because when we just solve for x, we get the same answer. Doing the quadratic we'd do: \[\frac{ -0 \pm \sqrt{0^2-4(1)(-2)} }{ 2(1) }\]

So now we have: \[\frac { \pm \sqrt {8} }{ 2 } = \frac{ 2 \pm \sqrt{2} }{ 2 }= \pm \sqrt{2}\]

Agreed?

Answer 13

But see we just did alll what grunt work when we could of done this: \[0=x^2-2\]
Add two to both sides:
\[2=x^2\]
Take the square root of both sides to isolate x
:\[\pm \sqrt{2}=x\]
And it's always plus/minus with square roots and finding x. Then boom, the same answer a faster way because it was possible to solve for 0 in this way.

Answer 14

@marissalovescats When I got to how did you get from +/- square root of 8 divided by 2 to 2 +/- square root of 2 divided by 2?

Answer 15

I simplified the square root of 8

Answer 16

\[\sqrt{8} = \sqrt{4 * 2} = 2\sqrt{2}\]

Because 4 is the highest perfect square you can multiply by another number to get 8, which is 4 * 2 and because 4 is a perfect square (2 times 2) it gets square rooted and we're left with a 2 outside of the radical.

Answer 17

I ended up with x = 1.4 because I took the approximate square root of 8 which was 2.8.

Answer 18

You usually don't want to do that because square roots are exact and not an approximate number

Answer 19

You should leave it expressed as radicals

Answer 20

@marissalovescats Oh, I see how you simplified that. The shortcut was really helpful too! I was hesitant to take an approximate because square roots are exact but I couldn't figure out what else to do! Thank you for all the help, I really appreciate it!!

Answer 21

Anytime! And remember even if you left it as rad8 you have rad8/2 which would still equal 1.8 and not 2.8

But you want to get into the habit of simplifying your radicals and leaving them in radicals, that's how they want it in higher math and it just makes things less messy without all those messy "approximate" numbers.

Answer 22

@marissalovescats Okay, thank you! & agreed, the approximates do get messy.

Answer 23

No problem, anytime!