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Question 1179118: A number exceed another by 5, the sum of their square is 157. What are the numbers?
Found 3 solutions by collegealgebraexpert, mananth, Theo:
Answer by collegealgebraexpert(1)   (Show Source):
You can put this solution on YOUR website! A number exceed another by 5, the sum of their square is 157. What are the numbers?
Solution:
Let the first number be x.
Then the second number will be x + 5.
Sum of the squares of the numbers = 157
x^2 + (x + 5)^2 = 157
x^2 + x^2 + 10x + 25 = 157
2x^2 + 10x + 25 - 157 = 0
2x^2 + 10x - 132 = 0
2(x^2 + 5x - 66) = 0
x^2 + 5x - 66 = 0/2
x^2 + 11x - 6x- 66 = 0
x(x + 11) - 6(x + 11) = 0
(x + 11)(x - 6) = 0
x + 11 = 0 or x - 6 = 0
x = -11 or x = 6
When we take x = -11, x + 5 becomes -11 + 5 = -6
When we take x = 6, x + 5 becomes 6 + 5 = 11
Thus, the two numbers are -11, -6 or 6, 11.
Answer by mananth(16946)  (Show Source):
You can put this solution on YOUR website! If x is one number the other is x+5
The sum of their squares is 157
x^2+(x+5)^2=157
x^2+x^2+10x+25=157
2x^2+10x-132=0
divide by 2
x^2+5x-66=0
x^2+11x-6x-66 =0
x(x+11)-6(x+11)=0
(x+11)(x-6)=0
x=-11, or x=6
when one number is -11 the is -6
when the number is 6 the othr is 11
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! let x = one of the numbers and y = the other.
if x exceeds y by 5, then x = y + 5
the sum of their squares is 157.
this means x^2 + y^2 = 157
since x = y + 5, then replace x with y + 5 in that equation to get:
(y + 5)^2 + y^2 = 157
simplify to get:
y^2 + 10y + 25 + y^2 = 157
combine like terms to get:
2y^2 + 10y + 25 = 157
subtract 157 from both sides of the equation to get:
2y^2 + 10y - 132 = 0
divide both sides of this equation by 2 to get:
y^2 + 5y - 66 = 0
factor this quadratic equation to get:
(y + 11) * (y - 6) = 0
solve for y to get:
y = -11 or y = 6
when y = 6, x = y + 5 = 11
when y = -11, x = y + 5 = -6
when y = 6 and x = 11, then 6^2 + 11^2 = 36 + 121 = 157
when y = -11 and x = -6, then (-11)^2 + (-6)^2 = 121 + 36 = 157.
your numbers can be:
(6 and 11) or (-11 and -6).
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