Question

Which equation has solutions of 6 and -6?

x^2 - 12x + 36 = 0
x^2 + 12x - 36 = 0
x²+36=0
x^2- 36 = 0

Answer 1

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Let's solve all of them to see :

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`{x}^(2) - 12x + 36 = 0`

`({x - 6})^(2 ) = 0`

`x - 6 = 0`

`x = 6`

There's only one solution , thus it's not what we want.

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`{x}^(2) + 12x - 36 = 0`

`(x - ( - 6 + 6 √(2)) \: )(x - ( - 6 - 6 √(2)) \: ) = 0 \\`

`x - ( - 6 + 6 √(2) ) = 0`

`x = - 6 + 6 √(2)`

And

`x - ( - 6 - 6 √(2) ) = 0`

`x = - 6 - 6 √(2)`

There are two solutions but none of them is what we want .

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`{x}^(2) + 36 = 0`

`{x}^(2) = - 36`

There's no solution because square of any number is greater than or equal zero

which means :

`{x}^(2) \geqslant 0`

Thus x² never can be a negative number.

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`{x}^(2) - 36 = 0`

`(x - 6)(x + 6) = 0`

`x - 6 = 0`

`x = 6`

And

`x + 6 = 0`

`x = - 6`

These are exactly the solutions what we want ;

Thus the correct answer is the last one.

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