divide (5x^2 - 6 X) ÷ 3x
pls solve this question
Answers
If we are given x^2=9 then can we write that, taking square root on both sides(not the principle square root only, ±x=±√9? So +x=±√9 or -x=±√9,then x=√9,x=-√9,-x=√9,-x=-√9?finally x=±3
It is not exactly like that . But the final thing can be accepted.
First of all , square roots are defined only when the inputs are positive (Square root of negative numbers is not defined in Real number system)
Square roots , always gives you the positive output , regardless of the number inside
For eg, (−9⋅−9)−−−−−−−−√=9 and not −9 .
So in square roots , there is not like principal square root or something like that.
So , in your example , x2=9 , when you take square root , the actual process happens is that
x2−−√=9–√
|x|=3
Here , LHS is known as the modulus of x . It is a type of function that will give you the magnitude of the given value . i.e only positive output.
For eg:
|−5|=5,|7|=7
|5|=5
In our case , |x|=3
Since , |3|=|−3|=3
We can say that , x=±3
It is understood that modulus is taken.
if x2=9 , then x=±9–√=±3
So , that’s how we remove the square root .
Answer:
(5x^2 - 6 X) ÷ 3x
=> (5x^2 ÷ 3x) - (6x ÷ 3x)
=> 5/3 x - 2