Asked by Raquel
Two numbers differ by 6. The sum of their squares is 116. Find the biggest number.
Answers
Answered by
Shenaya
Let the bigger number b "y" and the other will be "y-6" as they differ by 6.
The sum of their squares=y^2+(y-6)^2=116
(y-6)^2=y^2 -2*6y + 6^2
So 2y^2 - 12y + 36 =116
Now solve for y.You can use quadratic formula.
The sum of their squares=y^2+(y-6)^2=116
(y-6)^2=y^2 -2*6y + 6^2
So 2y^2 - 12y + 36 =116
Now solve for y.You can use quadratic formula.
Answered by
Henry
x - y = 6.
X = y + 6
x^2 + y^2 = 116.
(y+6)^2 + y^2 = 116
y^2 + 12y + 36 + y^2 = 116
2y^2 + 12y + 36 = 116
Divide by 2:
y^2 + 6y + 18 = 58.
y^2 + 6y - 40 = 0.
(y-4)(y+10) = 0
y-4 = 0, Y = 4.
y+10 = 0, Y = -10.
::
Use the positive value of Y: Y = 4.
x-y = 6
x - 4 = 6
X = 10.
X = y + 6
x^2 + y^2 = 116.
(y+6)^2 + y^2 = 116
y^2 + 12y + 36 + y^2 = 116
2y^2 + 12y + 36 = 116
Divide by 2:
y^2 + 6y + 18 = 58.
y^2 + 6y - 40 = 0.
(y-4)(y+10) = 0
y-4 = 0, Y = 4.
y+10 = 0, Y = -10.
::
Use the positive value of Y: Y = 4.
x-y = 6
x - 4 = 6
X = 10.
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