The coordinates of A and B are 27 square root and 15 square root respectively. Find AB.
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briefcaseTeacher (K-12)
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Find AB if the coordinate of A is `sqrt(27)` and the coordinate of B is `sqrt(15)` :
The distance d between two points on a number line a,b is `d=|a-b|` .
So `AB=|sqrt(27)-sqrt(15)|` . Since `sqrt(27)>sqrt(15)` we can drop the absolute value bars.
Then `AB=sqrt(27)-sqrt(15)` . This can be simplified:
`sqrt(27)=sqrt(9*3)=sqrt(9)sqrt(3)=3sqrt(3)`
`sqrt(15)=sqrt(5*3)=sqrt(5)sqrt(3)`
`AB=sqrt(27)-sqrt(15)=3sqrt(3)-sqrt(5)sqrt(3)`
`=sqrt(3)(3-sqrt(5))`
`~~1.323`
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Firstly we see that neither 27 nor 15 is a perfect square,
Perfect squares are numbers we get by multiplying one number with itself.
Eg-: 4(.i.e 2×2),9(.i.e 3×3),81(.i.e 9×9),etc
Here,we can write,
27=3×3×3 and
15=3×5
Where,A=square root 27
And B=square root 15.
Thus,AB =Square root 27× square root 15
= Square root(3×3×3) × square root(5×3)
=3×square root3×square root 3×square root 5
= 3×3×square root5
=3×3×2.2360
=20.12454(approx)
*we notice that while square rooting,the basic we follow is two of the same number is taken singly after being square rooted.
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