## Saturday, August 5, 2017

### Algebra: Solving sqrt(x + 15) = sqrt(15) – sqrt(x)

Algebra:  Solving sqrt(x + 15) = sqrt(15) – sqrt(x)

Introduction

You might have seen an ad that asks the seer to solve the following equation lately (if my memory serves me correctly):

√(x + 15) = √15 - √x

Solving √(x + 15) = √15 - √x

If it fairly easy to solve this equation for x.  First is to square both sides:

√(x + 15) = √15 - √x
(√(x + 15))^2 = (√15 - √x)^2
x + 15 = (√15)^2 – 2*√15*√x + (√x)^2
x + 15 = 15 – 2*√15*√x + x

Note that we can subtract x + 15 from both sides to obtain:
0 = -2*√15*√x

Divide by -2*√15 to get:
0 = √x

Squaring both sides (obviously, 0^2 = 0 * 0 = 0):
0 = x

Hence our solution is x=0.

But how about √(x + 15) = √15 + √x?

Solving √(x + 15) = √15 + √x

√(x + 15) = √15 + √x
(√(x + 15))^2 = (√15 + √x)^2
x + 15 = (√15)^2 + 2*√15*√x + (√x)^2
x + 15 = 15 + 2*√15*√x + x
0 = 2*√15*√x
0 = √x
0 = x

Like the previous equation, the solution is x=0.

The General Case

Using the techniques above we can easily see that the solutions to both:

√(x + A) = √A - √x
√(x - A) = √A + √x

Is x = 0 for any A.  Let’s assume that A is positive and nonzero.

√(x + A) = √A ± √x
(√(x + A))^2 = (√A ± √x)^2
x + A = (A)^2 ± 2*√15*√x + (√x)^2
x + A = A ± 2*√A*√x + x
0 = ± 2*√A*√x
0 = √x
0 = x

Note the properties used:

(x – y)^2 = x^2 – 2*x*y + y^2
(x + y)^2 = x^2 + 2*x*y + y^2
(√(x ± y))^2 = x ± y
And (± x)^2 = x^2 since x * x = x^2 and –x * -x = x^2

Now let’s tackle the cases √(x - 15) = √15 - √x and √(x - 15) = √15 + √x.  What do you think the solutions are?

Solving √(x - 15) = √15 - √x

√(x - 15) = √15 - √x
(√(x – 15))^2 = (√15 - √x)^2
x – 15 = (√15)^2 – 2*√15*√x + (√x)^2
x – 15 = 15 – 2*√15*√x + x

Subtracting x – 15 from both sides yields:

-30 = -2*√15*√x

Divide by 2*√15 to get (and note that –x/-y = x/y):

30/(2*√15) = √x

Squaring both sides gets:

900/(4*15) = x
900/60 = x
15 = x

Solving √(x - 15) = √15 + √x

√(x - 15) = √15 + √x
(√(x – 15))^2 = (√15 + √x)^2
x – 15 = (√15)^2 + 2*√15*√x + (√x)^2
x – 15 = 15 + 2*√15*√x + x
-30 = 2*√15*√x
-30/(2*√15) = √x

Squaring both sides gets:

900/(4*15) = x
15 = x

Can we infer what the solution for the general case is?

General Case

√(x - A) = √A ± √x
(√(x - A))^2 = (√A ± √x)^2
x - A = (A)^2 ± 2*√15*√x + (√x)^2
x - A = A ± 2*√A*√x + x

Here are going to break this into two cases:

Case 1:
x - A = A + 2*√A*√x + x
-2*A = 2*√A*√x
-(2*A)/(2*√A) = √x
-A/√A = √x
(-A/√A)^2 = (√x)^2
A^2/A = x
A = x

Case 2:
x - A = A - 2*√A*√x + x
2*A = 2*√A*√x
(2*A)/(2*√A) = √x
A/√A = √x
A^2/A = x
A = x

Either way, x = A

Eddie

This blog is property of Edward Shore, 2017