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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${x}^{2} + 12x - 36 = 0$

$(x - ( - 6 + 6 \sqrt{2}) \: )(x - ( - 6 - 6 \sqrt{2}) \: ) = 0$

$x - ( - 6 + 6 \sqrt{2} ) = 0$

$x = - 6 + 6 \sqrt{2}$

And

$x - ( - 6 - 6 \sqrt{2} ) = 0$

$x = - 6 - 6 \sqrt{2}$

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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