I've seen problems like this before on RUclips. The symmetry of the problem suggests that x and y are conjugates, i.e. x=a+b and y=a-b. Trying that: x+y = a+b+a-b = 2a = 8 So: a = 4 Then: xy = a^2 - b^2 = 16 - b^2 = 48 Rearranging and simplifying: b^2 = -32 So: b = ±4i√2 Then: x = a+b = 4+4i√2 or 4-4i√2 And: y = 8-x = 4-4i√2 or 4+4i√2 I enjoy your interesting problems.
With Vieta, we know, that those x and y in (x+y=-b=8 and x*y=c=48) are just the solutions of z^2 + bz + c = 0, so you immediately get 4:14 z^2-8z+48=0 (x or z doesn't matter, I didn't want to reuse x!) => z_12= 4 +/- 4 sqrt(2) i. Thus x_1 = z_1 and y_1 = z_2 || x_2 = z_2 and y_2 = z_1, that is x is one of the two z solutions and y is the other one.
x + y = 8
xy = 48
x? y?
=========
x + y = 8
x = 8-y
(x+y)² = x² + y² + 2xy
8² = (8-y)² + y² + 2(48)
64 = 64 - 16y + 2y² + 96
0 = 2y² - 16y + 96
0 = y² - 8y + 48
D = (-8)² - 4*1*48
D = -128 --> 2 imaginary roots 😅😅😅
I dislike the imaginary roots 😢😢😢
I've seen problems like this before on RUclips. The symmetry of the problem suggests that x and y are conjugates, i.e. x=a+b and y=a-b.
Trying that: x+y = a+b+a-b = 2a = 8
So: a = 4
Then: xy = a^2 - b^2 = 16 - b^2 = 48
Rearranging and simplifying: b^2 = -32
So: b = ±4i√2
Then: x = a+b = 4+4i√2 or 4-4i√2
And: y = 8-x = 4-4i√2 or 4+4i√2
I enjoy your interesting problems.
Just move y to the other side x=8-y substitute it into xy=48 and solve, then solve for x after done y, basic algebra.
@TidakTerdefinisi wow that was a lot of effort to show all the work nice job!
Binomial theorem 💀
With Vieta, we know, that those x and y in (x+y=-b=8 and x*y=c=48) are just the solutions of z^2 + bz + c = 0,
so you immediately get 4:14 z^2-8z+48=0 (x or z doesn't matter, I didn't want to reuse x!) => z_12= 4 +/- 4 sqrt(2) i.
Thus x_1 = z_1 and y_1 = z_2 || x_2 = z_2 and y_2 = z_1, that is x is one of the two z solutions and y is the other one.
by vietta this is equivalent to solving z^2-8z+48. what is lil bro doing for 10 mins 😂😂😂
Messy...
x+y=8. Multiply through by x => x^2 +xy=8x
Then, substituting xy=48 gives x^2+48=8x.
Re-arranging => x^2-8x+48=0
x+y=8
xy=48 -----> y=(48/x)
x+(48/x)=8
x²+48=8x
x²-8x+48=0
x²-8x+16=-32
(x-4)²=-32
|x-4|=4i√2
x-4=4i√2
x=4+4i√2 ❤
4+4i√2+y=8
4i√2+y=4
y=4-4i√2 ❤
x-4=-4i√2
x=4-4i√2 ❤
4-4i√2+y=8
-4i√2+y=4
y=4+4i√2 ❤
Also can multiply both sides of 1st eq. By y, substitute xy by 48 and solve for y.
Good review. Thanks!
Here is no real solution! y=4±4√2i and x=48/4±4√2i that would be the complex solution.
You only have to square the first equation and times two the second equation
(xy ➖ 2xy+1) .(xy ➖ 3xy+2).
8 2^3 2^1 (xy ➖ 2xy+1) 48 6^8 3^22^3 3^1^2^1 3^2 (xy ➖ 3xy+2)
Using x=8cos^2t And y=8sin^2t
Just use substitution method. This is std 10 maths
Painfully slow solution. Obviously x and y are interchangeable, so the 2 solutions are identical, just permuted.
Basic algebra.
One who have learnt 6 th well then he
Just use the quadratic formula lol
No real solutions exist
y=8-x => x(8-x)=48..........either x=12 or x=-4
No or itll be -48 for both not 48 thats why its not a real solution
x^2 - 8x + 48 = 0 to the quadratic equation lol
Or ... (and maybe shorter) ...
| x + y = 8
Hiup😅😅😅
Ridicule!!!