Sum of two numbers=68. The square of the larger less four times the square of the smaller=1204. Numbers are?

Question

What exactly are the two numbers?

Also please explain the solution.

Answer 1

x = larger number
y = smaller number

x + y = 68
x^2 - 4y^2 = 1204

y = 68 - x

x^2 - 4(68 - x)^2 = 1204
x^2 - 4(4624 - 136x + x^2) = 1204
x^2 - 4x^2 + 544x - 18496 = 1204
-3x^2 + 544x - 18496 = 1204
-3x^2 + 544x - 19700 = 0
3x^2 - 544x + 19700 = 0
(3x - 394)(x - 50) = 0
x = 50, 131 1/3

There are two answer sets:

(50, 18) and (131 1/3, -63 1/3)

Answer 2

We have two numbers. Lets call the number A and number B.
We know that...
A + B = 68

And one of these is true...

A ^2 - 4(B^2) = 1204

B ^2 - 4(A^2) = 1204

By moving around the first equation we can make...

A = 68 - B
B = 68 - A

You want to make one of the above equations contain all the same variables So used those two to substitute

(68 - B)^2 - 4(B^2) = 1204

(68 - B)(68 - B) - 4 B ^2 = 1204
4624 -68B -68B - B^2 -4B^2 = 1204
4624 -136B -5B^2 = 1204

Hope that gets you started...

Answer 3

The two numbers are 50 and 18.

50 squared is 2500
18 squared is 324, times 4 is 1296

2500 minus 1296 is 1204

Answer 4

x + y = 68
x^2 - 4y^2 = 1204

x = -y + 68

(-y + 68)^2 - 4y^2 = 1204
(-y + 68)(-y + 68) - 4y^2 = 1204
(y^2 - 68y - 68y + 4624) - 4y^2 = 1204
y^2 - 136y + 4624 - 4y^2 = 1204
(1 - 4)y^2 - 136y + 3420 = 0
-3y^2 - 136y + 3420 = 0
-(3y^2 + 136y - 3420) = 0
-(y - 18)(3y + 190) = 0

y = 18 or (-190/3)

Since you wanted a number and not a fraction of a number

y = 18

x + y = 68
x + 18 = 68
x = 50

The numbers are 50 and 18

Answer 5

As dscot399 said, the numbers are 50 and 18.

Answer 6

68 and 54

Answer 7

50 and 18. i am sure. do it by assuming one variable as 'x' and other as 'y'. you will get two equations in x and y. solve it

Answer 8

50 and 18 is correct. Two answers in two unknowns. Thanks!