**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

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