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The sum of the squares of two consecutive real numbers is 61. Find the numbers.

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blueglove | Student, Grade 11 | eNoter

Posted November 22, 2010 at 2:12 AM via web

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The sum of the squares of two consecutive real numbers is 61. Find the numbers.

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giorgiana1976 | College Teacher | Valedictorian

Posted November 22, 2010 at 2:12 AM (Answer #1)

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Let x and x+1 be the two consecutive numbers. We know, from enunciation that the sum of their squares is 61:

x^2 + (x+1)^2 = 61

We'll expand the square using the formula:

(a+b)^2 = a^2 + 2ab + b^2

We'll put a = x and b = 1:

x^2 + x^2 + 2x + 1 = 61

We'll combine like terms and we'll get:

2x^2 + 2x - 60 = 0

We'll divide by 2:

x^2 + x - 30 = 0

We'll apply the quadratic formula:

x1 = [-1+sqrt(1 + 120)]/2

x1 = (-1+11)/2

x1 = 5

x2 = -6

The first solution for the given problem is:

x = 5 => x+1 = 5+1 = 6

The 2 consecutive numbers whose sum of the squares is 61 are: 5,6.

5^2 + 6^2 = 25 + 36 = 61

The secondsolution for the given problem is:

x = -6 => x+1 = -6+1 = -5

The 2 consecutive numbers whose sum of the squares is 61 are: -6,-5.

(-6)^2 + (-5)^2= 61

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted November 22, 2010 at 2:14 AM (Answer #2)

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Let the first number be x

Then the next number will be x + 1.

Given that the sum of the squares  is 61

Then :

x^2 + ( x+1)^2 = 61

==> x^2 + x^2 + 2x + 1 = 61

We will combine like terms:

==> 2x^2 + 2x + 1 - 61 = 0

==> 2x^2 + 2x - 60 = 0

\We will divide by 2:

==> x^2 + x - 30 = 0

Now we will factor:

==> ( x - 6)( x+ 5)

==> x 1= 6  ==>  the second number = 5

==> x2= -5 ==> the second number = -6

Then the numbers are:

( -5, -6 )  OR  ( 5, 6)

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neela | High School Teacher | Valedictorian

Posted November 22, 2010 at 2:21 AM (Answer #3)

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We take x and x+1 as two consecutive numbers.

Then x^2+(x+1)^2 = 61, as the sum the squares is given to be 61.

So x^2 + x^2+2x+1 = 61.

We simplify the above as a quadratic equation ax^2+bx+c = 0.

2x^2+2x+1- 61 = 0.

2x^2+2x-60 = 0.

We divide both sides by 2:

x^2+x-30 = 0.

x^2+6x-5x -30 = 0.

x(x+6) - 5(x+6) = 0.

(x+6) (x-5) = 0.

We equate the factors to zero.

x+6 = 0 gives: x= -6.

x-5 = 0 gives: x= 5.

So x= 5 and x= 6 are the two consecutive numbers whose squares are 25 and 36. And sum of the squares is 25+36 = 51.

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted November 22, 2010 at 2:16 AM (Answer #4)

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We are given that the sum of the squares of two consecutive numbers is equal to 61. Let the numbers be x and x+1.

So x^2 + (x+1)^2 = 61

=> x^2 + x^2 + 1 + 2x = 61

=> 2x^2 + 2x - 60 = 0

=> x^2 + x - 30 =0

=> x^2 + 6x - 5x - 30 =0

=> x(x+6) -5(x+6) =0

=>(x-5)*(x+6) =0

We can have x = 5 and x = -6

Therefore the numbers can be 5 and 6 or -6 and -5.

The required result is (5, 6)  and (-6,-5).

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