the equation (x + a)(x - a) = 16

shahar

shahar

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There is question what is the value of a in the equation:
(x + a)(x -a) = 16

a) 4
...
another answers...
...
d) Neither of above choices

The answer is d. The answer is can be 4 or -4. so the answer d is the correct.
Can be another answers to a in this equation?
Is there a proof that they are the only answers (a = 4, a = -4)?
 
There is no answer since you consider two variables. The solutions to this equation are many pairs [imath] (x,a). [/imath] This is in a mathematical notation
[math] \left\{(x,a)\in \mathbb{R}^2\,|\,x^2=16+a^2\right\}\,. [/math]There are many possible combinations. E.g., [imath] a=0 [/imath] means that [imath] x=\pm 4, [/imath] and [imath] a=\pm 3 [/imath] means that [imath] x=\pm 5. [/imath]
 
shahar said:
There is question what is the value of a in the equation:
(x + a)(x -a) = 16

a) 4
...
another answers...
...
d) Neither of above choices

The answer is d. The answer is can be 4 or -4. so the answer d is the correct.
Can be another answers to a in this equation?
Is there a proof that they are the only answers (a = 4, a = -4)?
Click to expand...
You can only say what a is if you first know what x is (or, perhaps, if you are told something about the equation). The question makes no sense as it stands.

Did you omit something? Please show the entire original.
 
Dr.Peterson said:
You can only say what a is if you first know what x is (or, perhaps, if you are told something about the equation). The question makes no sense as it stands.

Did you omit something? Please show the entire original.
Click to expand...
The correct equation (4 - a)(4 + a) = 0
I make mistake.
The 2 values of a is -4 or 4 like in the opening post.
So why there are 2 values and there are no more?
 
The easiest way to see it is the observation that a product can only be zero if one of the factors is. So we have either [imath] 4-a= 0[/imath] which means [imath] a=4 [/imath] or [imath] 4+a=0 [/imath] which means [imath] a=-4. [/imath]

There are no other ways for a product to be zero.
 
shahar said:
you say: [There are no other ways for a product to be zero.]

Why? Is there a proof to this statement?
Click to expand...

Is it not clear that the product of two nonzero numbers is nonzero? So if a product ab = 0, then either a or b must be zero.

You can prove it formally if you want: Suppose that ab = 0, and a is not zero. Then b = ab/a = 0/a = 0. QED.
 
shahar said:
you say:

Why? Is there a proof to this statement?
Click to expand...
This depends on where you want to start from. E.g., if you accept that any finite sum of ones won't ever be zero, then we can conclude that if [math] a\cdot b=0=\underbrace{(1+\ldots+1)}_{a\text{ times}}\cdot \underbrace{(1+\ldots+1)}_{b\text{ times}}=\underbrace{1+\ldots+1}_{a\cdot b\text{ times}} [/math]cannot be zero unless [imath] a [/imath] or [imath] b [/imath] is. That [imath] 0\cdot a=0 [/imath] follows from the distributive law:
[math] a\cdot 0=a\cdot (1-1)=a\cdot 1 - a\cdot 1=a-a=0.[/math]
That any sum of ones isn't zero is not always true. A computer (or a light switch) only knows on and off, and on+on=off. The small hand of a clock repeat after [imath] 12 [/imath] steps, so [imath] 1+1+1+1+1+1+1+1+1+1+1+1=0. [/imath] But the integers are made so that any sum of ones is never zero. This follows from the construction of natural numbers: every natural number has a new follower, so we cannot return to zero.
 
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(1, 16), (2, 8), (4, 4)

Also, [imath]x^2 - a^2 = 16 \implies x^2 = a^2 + 4^2 \implies 5^2 = 3^2 + 4^2 \implies x = 5, a = 3[/imath]
 
Agent Smith said:
(1, 16), (2, 8), (4, 4)

Also, [imath]x^2 - a^2 = 16 \implies x^2 = a^2 + 4^2 \implies 5^2 = 3^2 + 4^2 \implies x = 5, a = 3[/imath]
Click to expand...
👏👏
I loved how you cracked it Smith. You are absolutely from another planet. You can see things from a completely different angle! Are you a professor or something?

🤔
 
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