# Thread: [SOLVED] Something bothering my understanding

1. ## [SOLVED] Something bothering my understanding

The upper part is the question and lower part is my answer.
Is a=-1, b=3, c=1 acceptable?

2. Originally Posted by stpmmaths

The upper part is the question and lower part is my answer.
Is a=-1, b=3, c=1 acceptable?
HI

Yes , its acceptable as one of the answers . You will still need to test it and see whether it works fine with the matrix .

$\displaystyle \sqrt{1} = \pm 1$ but in this case is -1 .

However , if $\displaystyle a=-1$ , then its not acceptable .

3. Originally Posted by stpmmaths

The upper part is the question and lower part is my answer.
Is a=-1, b=3, c=1 acceptable?
Have you substituted a = 1, b = 3 and c = -1 into the matrix? Do you get a symmetric matrix?

4. If I substituted a=1, b=3 and c=-1, I can't get a symmetric matrix. However by taking $\displaystyle \sqrt{a}$ = -1, I can get a symmetric matrix.

5. $\displaystyle \sqrt{a}$ = -1
By squaring both side,
a = 1
Is this true?

But when we substitute back a = 1 into $\displaystyle \sqrt{1}$ is not equal to -1.

How?

HI

Yes , its acceptable as one of the answers . You will still need to test it and see whether it works fine with the matrix .

$\displaystyle \sqrt{1} = \pm 1$ but in this case is -1 .

However , if $\displaystyle a=-1$ , then its not acceptable .
I don't understand this equation $\displaystyle \sqrt{1} = \pm 1$. I thought it should be $\displaystyle \pm \sqrt{1} = \pm 1$

7. $\displaystyle \sqrt{1}= \pm 1$ is incorrect. $\displaystyle \sqrt{1}= 1$ since "$\displaystyle \sqrt{a}$" is defined as "the positive x such that $\displaystyle x^2= a$".

It is true that the roots of the equation $\displaystyle x^2= 1$ are 1 and -1, but the reason we say "the roots of equation $\displaystyle x^2= a$ are $\displaystyle x= \pm\sqrt{a}$ is that "$\displaystyle x= \sqrt{a}$" does NOT give both roots.

8. Originally Posted by stpmmaths
I don't understand this equation $\displaystyle \sqrt{1} = \pm 1$. I thought it should be $\displaystyle \pm \sqrt{1} = \pm 1$
gosh , i am wrong again . Sorry stpm math , Hallsofivy has corrected me .

9. Originally Posted by HallsofIvy
$\displaystyle \sqrt{1}= \pm 1$ is incorrect. $\displaystyle \sqrt{1}= 1$ since "$\displaystyle \sqrt{a}$" is defined as "the positive x such that $\displaystyle x^2= a$".

It is true that the roots of the equation $\displaystyle x^2= 1$ are 1 and -1, but the reason we say "the roots of equation $\displaystyle x^2= a$ are $\displaystyle x= \pm\sqrt{a}$ is that "$\displaystyle x= \sqrt{a}$" does NOT give both roots.
So in this case, $\displaystyle \sqrt{a} = -1$ does not exist. Am I right?

10. It is incorrect! I'm not sure I would say "it doesn't exist". That equation certainly does exist- it's right there on the screen!

And both $\displaystyle \sqrt{a}$ and -1 exist. They just can't be equal!

11. So a=1, b=3 and c=-1 is not the answer. Right?

12. ## Precisions about the question

Hello,
By finding the notation ($\displaystyle \sqrt{a}$) in the text of your question, it is certain that your exercice concern the set of real numbers (not complex numbers) .
The equality $\displaystyle c = \sqrt{a}$ makes sence if and only if the variables a and c are defined as positive real numbers (includes the possible value of zero).
So finding c=-1 makes the value of c out of defined range and thus there is no need to continue calculating the values of a and b.