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