if the difference of 2 numbers is 10 and their product is 12, what is the sum of their squares?

Question

please help

Answer 1

The first step is to find the two numbers, x and x+10

x * (x+10) = 12

x^2 + 10 x - 12 = 0
a quadratic
(there may be more than one answer)

Go for it.

---

x * (x+10) = 12

(x+10) = 12/x

x^2 + (x+10)^2 = x^2 +144/x^2

sum of squares = (x^4 + 144) / x^2

PS:
Come on, benoit3535, my radicals are not ungodly. They may be messy... But at least, they are real.

:-)

(for those who have not seen this: it is a smile -- I am joking)

x = (-10 +/- SQRT(100 + 48))/2

x = -5 +/- SQRT(37)

take the + version

x = -5 +√37 x+10 = +5 + √37

x*(x+10) = (-5+√37)*(+5+√37) =

-25 -5√37 +5√37 + 37 = 12

Sum of squares:

(-5+√37)^2 + (+5+√37)^2 = 124

---

In the - version, the answer is the same
(shown by symmetry)

(-5-√37)^2 = (+5+√37)^2
and
(+5-√37)^2=(-5+√37)^2

---

and by now, you should know that I like to do things the "long way"...
Still, I voted for yours.

Answer 2

telkwa had the splendid concept 4 - d , 4 , 4 + d (the place d is the uncomplicated distinction) ***edit*** in case you're actually not particular the place 4 comes from, permit the middle quantity be x the 1st term is x - d and the 0.33 term is x + d (the place d is the uncomplicated differene) (x - d) + x + (x + d) = 12 3x = 12 x = 4 subsequently, the middle quantity must be 4 be conscious that that's oftentimes much less stressful to apply the middle quantity simply by fact the unknown, particularly than the smallest (or greatest) **** end edit **** product is 4(4 - d)(4 + d) = 80 sixteen - d^2 = 20 -4 = d^2 d = +/- 2i (the place i^2 = -a million) subsequently, the sequence (applying complicated numbers) is 4 - 2i , 4 , 4 + 2i examine: 4 - 2i + 4 + 4 + 2i = 12 (4 - 2i)(4)(4 + 2i) = 4(sixteen - 4i^2) = 4(20) = 80 so the words are 4 - 2i, 4, and four + 2i

Answer 3

x-y=10
x*y= 12.
squaring both sides of the first x^2+y^2 -2x*y =100
so x^2+y^2 = 100+2x*y = 124 The sum of their squares is 124

Answer 4

You can do as Raymond suggests, solve the quadratic equation, square the solutions and add them up which involves dealing with ungodly radicals... messy stuff.

Or you can take the lazy route which is this:

a-b = 10
ab = 12

→ a² + b² = (a-b)² + 2ab = 100 + 2*12 = 124.

This solution was brought to you by the equality
a² - 2ab + b² = (a-b)².

P.S.: Raymond, that's an irrational way of solving it, to say the least... ;-)

Answer 5

a - b = 10
ab = 12

find a²+b²

a = (b+10)
b(b+10)=12
b²+10b=12
b² = 12-10b

b=a-10
a(a-10)=12
a²-10a=12
a²=12+10a

a²+b²=12-10b + 12 + 10a
a²+b²=24 + 10(a-b)
a²+b²=24 + 10 (10)
a²+b² = 124