shubhim20
could you please explain this in detail
PyjamaScientist
\(\sqrt{9} = 3\\
\)
\(9^\frac{1}{3} = 2.x\). Since, \(9^\frac{1}{3}\) is greater than \(8^\frac{1}{3}\)
\(9^\frac{1}{4}\) would be greater than 1. Since, \(9^\frac{1}{4}\) is greater than \(1^\frac{1}{4}\)
So, 3 - 2.x - 1.y => 0.x - 1y. The result has to be negative. Thus < 0.
\(P = 9^{1/2} - 9^{1/3} - 9^{1/4}\)
\(9^\frac{1}{2} = 3\)
Now we are subtracting two values from \(3\), viz. \(9^{1/3}\) and \(9^{1/4}\)
\(9^{1/3}\) => it is difficult to calculate this, so we try to approximate a value. We know that
cube root of 8 is 2, therefore
cube root of 9 must be greater than 2Again, \(9^{1/4}\) => difficult to calculate fourth root, so we approximate.
Fourth root of 1 is 1, so fourth root of 9 must be greater than 1Now, to find P,
we will subtract something greater than 2, and something greater than 1 from 3P = 3 - something greater than 2 - something greater than 1
P = 3 - (something greater than 2 + something greater than 1)
P = 3 - (something greater than 3) =>
P is negativeHope it is clear.