# Can someone help me solve this quadratic

• 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

## 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?

What is (a+b)^2 equal to?
I am not sure what you mean?

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

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

this is a frequently made mistake.

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

this is a frequently made mistake.

Well (a+b)^2 = (a+b)(a+b) = ...
Use distributivity
Ahhhh I see, thank you!

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)
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

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

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.

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.