# Thread: how to solve by square root principle

1. ## how to solve by square root principle

Solve by using square root principle

[ (x+1)/2 ]^2 - (x/2) = 2

I tried the following:

[ (x+1)/2 ]^2 - (x/2) = 2

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

-bring the (1/4) to RHS

(x^2/4) - (x/2) = (7/4)

-LCD 4

x^2 - 2x = 7

I'm stuck now as I have a middle term (-2x), otherwise the square root principle would work out nice. How to solve this using the square root principle?

Thanks.

2. Originally Posted by shenton
Solve by using square root principle

[ (x+1)/2 ]^2 - (x/2) = 2

I tried the following:

[ (x+1)/2 ]^2 - (x/2) = 2

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

-bring the (1/4) to RHS

(x^2/4) - (x/2) = (7/4)

-LCD 4

x^2 - 2x = 7

I'm stuck now as I have a middle term (-2x), otherwise the square root principle would work out nice. How to solve this using the square root principle?

Thanks.
$[ (x+1)/2 ]^2 - (x/2) = 2$

expand the square:

$[ x^2/4+x/2+1/4 ]- (x/2) = 2$

$x^2+1=8$

$x^2=7$.

RonL

3. Thanks!

This part is confusing:

[ (x/2) + (1/2) ]^2

I worked it out as:

(x^2/4) + (x/4) + (x/4) + (1/4)

Add (x/4) with (x/4), this gives me (2x/4):

(x^2/4) + (2x/4) + (1/4)

Simplify (2x/4) by dividing by 2, this gives me (x/2):

(x^2/4) + (x/2) + (1/4)

Are the above workings right? Is that how I expand the binomial fraction?

Thanks.

4. Originally Posted by shenton
Thanks!

This part is confusing:

[ (x/2) + (1/2) ]^2

I worked it out as:

(x^2/4) + (x/4) + (x/4) + (1/4)

Add (x/4) with (x/4), this gives me (2x/4):

(x^2/4) + (2x/4) + (1/4)

Simplify (2x/4) by dividing by 2, this gives me (x/2):

(x^2/4) + (x/2) + (1/4)

Are the above workings right? Is that how I expand the binomial fraction?

Thanks.
Yes, you have expand it ok.

5. Thanks for confirming the workings.