 ##### sr
Mr Sunthorn SR Rathmanus

### โจทย์เลข x+y = 19, xy = 9, x√y+y√x = ?

It's a Sunday. Let's take it easy and relax - physically. But it's a day for reflection, recollection, review and replanning our activities – changing our life.

But many of us are happy with what we have done and what we are. The plan we have is working well and should be kept going – cruising along.

A little math quiz may help us look at many ways we solve problems in our life. This quiz come from Quora.com (which publishes a large collection of good questions and excellent answers in many fields of inquiry). The goal is not only to find the answer but also to find the easiest way to work out it out.

Given  x+y = 19 and xy = 9, find  x√y+y√x = ?

Please share your solution method so we can all learn more ways and become more efficient in solving problems together, if this ‘among gifts, the gift of knowledge is most valuable’ is not enough.

[Edited on 7 jun 2565BE, to include some solution paths below.]

A. By substituting (either x or y)

y = 19-x; then x(19-x) = 9; or 19x-x² = 9;
or a quadratic equation: x²-19x + 9 = 0
using the formula x = (-b±√(b²-4ac))/(2a) we get
y = -(-19)±√(19²-4*1*9)/(2*1) = 19 ±√(19²-6²)/2 --as 4*1*9 = 2²*3² = 6²
using a²-b²=(a+b)(a-b) we have √(19²-6²) = √((19-6)(19+6)) = √(13*25);
and using = √(ab) = √a√b we get √(13*25) = 5√13
to get  y = -(-19)±√(19²-4*1*9)/(2*1) = (19±5√13)/2
Here we should see that we hit a snag (√13) and hard work to go on (this solution path) to calculate the value of x√y+y√x.
x = 19-y;then y(19-y) = 9; or 19y-y² = 9 then we follow the same path as for y we will get x = (19±5√13)/2. Then we have to calculate √y, √x and x√y+y√x.)

B. By working back from 'x√y + y√x'

B.1 Multiply the given facts
(x+y)*(xy) = 19*9 -> x²y + xy² = 19*9
Add 2(x√y*y√x) to both LH and RH sides
x²y + 2(x√y*y√x) + xy² = 19*9 + 2(x√y*y√x)
using (a + b)² = a² + 2ab + b² and assume x, y are positive so √x√y = √(xy) proof?
(x√x+y√y)² = 19*9 + 2(xy*√x√y) = 19*9 +2(xy*√(xy))
Substitute xy=9 and calculate (ignore the negative value)
(x√x+y√y)²  = 19*9 + 2(9*√9) = 9(19+ 2√9) = 9*(19+6)= 9*25 = 3²*5² = (3*5)²
(x√x+y√y)  = 15 as required

B.2 Taking the square root of xy = 9  so √(xy) = √9 = 3
Rewriting (x + y) as ((√x)² + (√y)²) and adding 2√(xy) to both sides to show
(√x)² + (√y)² + 2√(xy) = 19 + 2√(xy)
using identity (a+b)² = a² +2ab + b² and substitute √(xy) = 3
(√x + √y)² = 19 + 2*3 = 25
√x + √y = ±5 or 5 --if we ignore negative values;
multiplying by both sides by √(xy) and using √(xy) = √x√y
√x√y√x + √x√y√y = 5*3 --> (x√y + y√x) = 15  as required

We can see that thinking differently can solve a (math) problem easier and quicker. It is not however easy to learn to think about 'other ways' or 'options', we need to ask ourselves and others 'how do you solve this problem?' instead of 'what is the answer?'.

And only practice will make us better problem solvers.

หมายเลขบันทึก: 702929เขียนเมื่อ 5 มิถุนายน 2022 05:47 น. ()แก้ไขเมื่อ 7 มิถุนายน 2022 17:55 น. ()สัญญาอนุญาต: สงวนสิทธิ์ทุกประการ

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