Can someone help me solve this quadratic !

  • Thread starter Tlark10
  • Start date
  • #1
4
1

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
 

Answers and Replies

  • #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?
 
  • #5
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
 
  • #6
4
1
this is a frequently made mistake.

Well (a+b)^2 = (a+b)(a+b) = ...
Use distributivity
Ahhhh I see, thank you!
 
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  • #7
34,688
6,394

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)##
Tlark10 said:
16x^2 +25 = 64 - 64x^2
 
  • #8
chwala
Gold Member
948
83
alternatively ##4x+5=8√(1-x)##, can be expressed as, ##(4x+5)^2=64(1-x)##
 
  • #9
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.
 
  • #10
chwala
Gold Member
948
83
ok elaborate please.
 
  • #11
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.
 
  • #12
34,688
6,394
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.
 
  • #13
chwala
Gold Member
948
83
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.
 

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