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√12 = 2√3
√12×√3 = 2√3×√3 = 2×3 = 6
√12×√3 = 2√3×√3 = 2×3 = 6
Answered by
5
ANSWER::
![\sqrt[2]{12} \times \sqrt{3 } \sqrt[2]{12} \times \sqrt{3 }](https://tex.z-dn.net/?f=+%5Csqrt%5B2%5D%7B12%7D+%5Ctimes+%5Csqrt%7B3+%7D+)
ACCORDING TO PRIME FACTORISATION:-
12= 2×2×3×1
3= 1×3
So,
![\sqrt[2]{12} \times \sqrt{3 } \sqrt[2]{12} \times \sqrt{3 }](https://tex.z-dn.net/?f=+%5Csqrt%5B2%5D%7B12%7D+%5Ctimes+%5Csqrt%7B3+%7D+)
=![\sqrt[2]{1×1×2×2×3×3} \sqrt[2]{1×1×2×2×3×3}](https://tex.z-dn.net/?f=+%5Csqrt%5B2%5D%7B1%C3%971%C3%972%C3%972%C3%973%C3%973%7D+)
In finding square root , always we have to take pair of numbers as single number which will be the square root of number.
So,
![\sqrt[2]{1×1×2×2×3×3} \sqrt[2]{1×1×2×2×3×3}](https://tex.z-dn.net/?f=+%5Csqrt%5B2%5D%7B1%C3%971%C3%972%C3%972%C3%973%C3%973%7D+)
= 1×2×3
= 6 (<-------ANSWER)
OR
![\sqrt[2]{1×1×2×2×3×3} \sqrt[2]{1×1×2×2×3×3}](https://tex.z-dn.net/?f=+%5Csqrt%5B2%5D%7B1%C3%971%C3%972%C3%972%C3%973%C3%973%7D+)
=![\sqrt[2]{36} \sqrt[2]{36}](https://tex.z-dn.net/?f=+%5Csqrt%5B2%5D%7B36%7D+)
= 6 (<------------ANSWER)
_________________________
✨✨BE BRAINLY ✨✨
ACCORDING TO PRIME FACTORISATION:-
12= 2×2×3×1
3= 1×3
So,
=
In finding square root , always we have to take pair of numbers as single number which will be the square root of number.
So,
= 1×2×3
= 6 (<-------ANSWER)
OR
=
= 6 (<------------ANSWER)
_________________________
✨✨BE BRAINLY ✨✨
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