# Thread: Modulus and Irrational Equations

1. ## Modulus and Irrational Equations

Hello all my name is larsson and I am 30 years of age and am trying to teach myself Maths. I was able to do most of the questions in this section but the last few have me stumped

(the square root of x) + 1 = ( the square root of x + 9 )

It says these questions need squaring twice and the answer is 16 but I can't work it out.

Any help would be great
Thanks and regards

2. $\sqrt{x} +1 = \sqrt{x+9}$

square both sides

$x + 2 \sqrt{x} +1 = x+9$
$2 \sqrt{x}= 8$
$\sqrt{x}= 4$
$x = 16$

With equations the involve squaring, it is important to check your solutions. This is because the process of squaring an equations is not reversible.

From going to your first step to your second step you introduce an additional equation into the problem.

The expression $x + 2 \sqrt{x} +1 = x+9$ could have either come form squaring this $\sqrt{x} +1 = \sqrt{x+9}$ or $- \sqrt{x} - 1 = \sqrt{x+9}$, so the solutions to obtain through this process could be solutions to either equation. additional solutions obtained form squaring an equation are called extraneous roots. In this case you don't have the problem but watch out for them.

Bobak

3. Thank you Bobak for taking the time to reply to me, just one thing I don't understand - how to get the second line of the solution

x + 2(square root of x ) + 1 = x + 9
Its the red part I don't understand

Thanks a million

square both sides

$x + 2 \sqrt{x} +1 = x+9$
$2 \sqrt{x}= 8$
$\sqrt{x}= 4$
$x = 16$

With equations the involve squaring, it is important to check your solutions. This is because the process of squaring an equations is not reversible.

From going to your first step to your second step you introduce an additional equation into the problem.

The expression $x + 2 \Sqrt{x} +1 = x+9$ could have either come form squaring this $\Sqrt{x} +1 = \Sqrt{x+9}$ or $- \Sqrt{x} - 1 = \Sqrt{x+9}$, so the solutions to obtain through this process could be solutions to either equation. additional solutions obtained form squaring an equation are called extraneous roots. In this case you don't have the problem but watch out for them.

Bobak[/quote]

I squared both sides.

we started with $\sqrt{x} +1 = \sqrt{x+9}
$

then

$(\sqrt{x} +1)^2 = (\sqrt{x+9})^2
$

$(\sqrt{x})^2 + 2 \sqrt{x} +1 = x+9$

$x + 2 \sqrt{x} +1 = x+9$

Is that more clear ?

Do you understand why this step is not reversible ?

5. Just to elaborate a tad more:

$(\sqrt{x} +1)^2 = (\sqrt{x} + 1) \cdot (\sqrt{x} + 1)$
$= \sqrt{x} \cdot \sqrt{x} + \sqrt{x} \cdot 1 + \sqrt{x} \cdot 1 + 1 \cdot 1$
$= (\sqrt{x})^2 + 2 \cdot \sqrt{x} + 1 = x + 2\sqrt{x} + 1$

6. ## Thanks

Thanks alot to Bobak and Flippy for taking the time to reply to me. I understand now how to do these equations.

Once again thank you both for spending time to help me understand!!!!

Regards