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If `x^2 - a = 14` and `x^2+a=16,` what is the value of `a` ?
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It is given that x^2 - a = 14.
This equation has two variables x and a. x^2 - a can be equal to 14 for infinite number of sets of x and a. It is not possible to determine a unique solution for a from the given equation.
As the equation x^2 - a = 14 has 2 independent variables it is not possible to determine a unique solution for a.
Posted by justaguide on September 27, 2013 at 2:50 PM (Answer #1)
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Sorry there is another equation x^2 + a = 16
Posted by timw996 on September 27, 2013 at 2:51 PM (Answer #2)
High School Teacher
I edited your original question in response to Answer #2.
If you subtract the first equation from the second one, you get `2a=2,` so `a=1.`
Althought it wasn't asked for, you can go further and solve for `x,` and we find that both `+-sqrt(15)` work. Finally, we can check our solution:
`(+-sqrt(15))^2-1=14` and `(+-sqrt(15))^2+1=16,` so it works.
Posted by degeneratecircle on September 27, 2013 at 3:03 PM (Answer #3)
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