23-02-2016
Mathematics

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23-02-2016

[tex] x^{2} =169\\\\\sqrt{x^2} =\pm\sqrt{169}\\\\x =\pm \sqrt{169}\\\\x =\pm 13\\\\\boxed{\bf{x = 13}}\\\boxed{\bf{x=-13}}[/tex]

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Billymkhoi Billymkhoi

23-02-2016

x^2= 169
⇒ x^2 -169= 0
⇒ x^2- 13^2= 0
⇒ (x-13)(x+13)= 0
⇒ x-13= 0 or x+ 13= 0
⇒ x= 13 or x= -13

The correct answer is 13 and -13.

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