# Modulus and Irrational Equations

• Apr 10th 2008, 03:42 AM
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
• Apr 10th 2008, 04:27 AM
bobak
$\sqrt{x} +1 = \sqrt{x+9}$

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
• Apr 10th 2008, 05:12 AM
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]
• Apr 13th 2008, 11:56 AM
bobak

Sorry for the late reply

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 ?
• Apr 13th 2008, 12:31 PM
Flippy
Just to elaborate a tad more:

$(\sqrt{x} +1)^2 = (\sqrt{x} + 1) \cdot (\sqrt{x} + 1)$
$= \sqrt{x} \cdot \sqrt{x} + \sqrt{x} \cdot 1 + \sqrt{x} \cdot 1 + 1 \cdot 1$
$= (\sqrt{x})^2 + 2 \cdot \sqrt{x} + 1 = x + 2\sqrt{x} + 1$
• Apr 15th 2008, 04:42 AM