Why Does 40x Appear in the Quadratic Equation Transformation?

[Tlark10]

Homework Statement

I can't determine where the 40x comes from.
From: 4x + 5 = 8√(1 - x)
To: 16x^2 +40x + 25 = 64 - 64x^2

Homework Equations

√1-x^2 = √1-x *√x+1

The Attempt at a Solution

4x + 5 = 8√(1-x)
4x^2 + 5^2 = (8 - 8x)*(8x + 8)
16x^2 +25 = 64 - 64x^2

 

[member 587159]

Homework Statement

I can't determine where the 40x comes from.
From: 4x + 5 = 8√(1 - x)
To: 16x^2 +40x + 25 = 64 - 64x^2

Homework Equations

√1-x^2 = √1-x *√x+1

The Attempt at a Solution

4x + 5 = 8√(1-x)
4x^2 + 5^2 = (8 - 8x)*(8x + 8)
16x^2 +25 = 64 - 64x^2

What is (a+b)^2 equal to?

 

[Tlark10]

What is (a+b)^2 equal to?

I am not sure what you mean?

 

[Tlark10]

What is (a+b)^2 equal to?

(a+b)^2 = a^2 + b^2

 

[member 587159]

(a+b)^2 = a^2 + b^2

this is a frequently made mistake.

Well (a+b)^2 = (a+b)(a+b) = ...
Use distributivity

 

[Tlark10]

this is a frequently made mistake.

Well (a+b)^2 = (a+b)(a+b) = ...
Use distributivity

Ahhhh I see, thank you!

 

[Mark44]

Homework Statement

I can't determine where the 40x comes from.
From: 4x + 5 = 8√(1 - x)
To: 16x^2 +40x + 25 = 64 - 64x^2

Homework Equations

√1-x^2 = √1-x *√x+1

The Attempt at a Solution

4x + 5 = 8√(1-x)
4x^2 + 5^2 = (8 - 8x)*(8x + 8)

What you wrote as a relevant equation doesn't apply here. $[8\sqrt{1 - x}]^2 \ne (8 - 8x)(8x + 8)$

16x^2 +25 = 64 - 64x^2

 

[chwala]

alternatively $4x+5=8√(1-x)$, can be expressed as, $(4x+5)^2=64(1-x)$

 

[member 587159]

alternatively $4x+5=8√(1-x)$, can be expressed as, $(4x+5)^2=64(1-x)$

One should be careful because those might not be equivalent.

 

[chwala]

ok elaborate please.

 

[member 587159]

ok elaborate please.

The first equation (the original one) has one solution for x. The second one has 2 solutions for x. We use indeed that => but to make sure you can use <=> you have to say that one solution for x is not valid.

 

[Mark44]

ok elaborate please.

The basic idea is that if you square both sides of an equation, the new equation might have solutions that don't satisfy the original equation. Here's a very simple example:

(Eqn 1) x = -2
Square both sides to get
(Eqn 2) $x^2 = 4$
The first equation has -2 as its solution. The second equation has -2 and 2 as its solutions.

 

[chwala]

agreed , we can have the two solutions as stated in post 12 ,but we can go ahead and state that one solution does not satisfy the equation, my take in post number 8, is to indicate that one way of solving such quadratics is by squaring both sides, then determine later which solutions satisfy the original equation.