One positive integer is 7 less than twice another. The sum of their squares is 145?

Answer 1

2x-7=z

z²+x²=145

(2x-7)²+x²=145

(2x-7)(2x-7)+x²=145

4x²-28x+49+x²=145

5x²-28x+49=145

5x²-28x-96=0

Now use the quadratic equation 

x=8

Now plugin x into the first equation to solve for z.

z=9

Answer 2

let the intger be  'n' & '2n - 7' 

Hence 

n^2 + ( 2n - 7)^2 = 145 

n^2 + 4n^2 - 28n + 49 = 145 

5n^2 - 28n - 96 = 0 

Factor 

(5n + 12)(n - 8) = 0 

n- 8 = 0 

n = 8 

Hence 

2n - 7 = 9 

Hence the integers are 8 & 9 

Answer 3

Let x & y be that 2 integers.

x+7=2y

x^2+y^2=145

=>

y^2+(2y-7)^2=145

=>

5y^2-28y-96=0

=>

(y-8)(y+12/5)=0

=>

y=8 or y=-12/5 (rejected)

x=2(8)-7=9

Thus, the integers are

8 & 9.

Answer 4

5n^2 - 28n - 96 = (5n + 12)(n - 8)

Only integer that works is 8

The sum of their squares is 145 = 64 + 81

9 is 7 less than twice 8

Answer 5

One positive integer is 7 less than twice another

x and 2x - 7

The sum of their squares is 145

x² + (2x - 7)² = 145

x² + 4x² - 14x -14 x + 49 = 145

5x² - 28x + 49 = 145

5x² - 28x -96 = 0

(5x + 12) (x - 8) = 0

5x = - 12 (reject as negative

x = 8

if x = 8

2x - 7 = 16 - 7 = 9

ANS 8 and 9

Answer 6

Let the two integers be x and y.

According to the given condition, 2 y - x = 7

=> x = 2 y - 7 .......... (1)

Second conditions says : sum of squares of x and y = 145,

=> x² + y² = 145 ..... (2)

from (1) and (2) we have ---

=> ( 2 y - 7 )² + y² = 145

=> 4 y² - 28 y + y² + 49 = 145

=> 5 y² - 28 y - 96 = 0

=> 5 y² - 40 y + 12 y - 96 = 0

=>  5y ( y - 8 ) + 12 ( y - 8 ) = 0

=> ( y - 8 ) ( 5 y + 12 ) = 0

=> y = 8 OR y =  - 12/5. Neglect fractional value since y is an iteger.

=> from (1)  x  =  9

Hence the two Integers are ( 8 and 9 ) ............... Answer

Answer 7

Since 13² > 145, we can just work back from 12 and there can't be more than 6 cases to  check.

145 - 12² = 1, which is square

(2 x 12) - 1 = 7 ✘ and (2 x 1) - 12 = 7 ✘

145 - 11² = 24 not square

145 - 10² = 45 not square

145 - 9² =  64 = 8²

8² + 9² = 145 ✓

(2 x 8) - 9 = 7 ✓

Answer 8

Excel

x   2x 2x-7  x^2 (2x-7)^2  Sum

1    2    -5    1    25            26

2    4    -3    4    9              13

3    6    -1    9    1              10

4    8    1    16    1              17

5    10    3    25    9             34

6    12    5    36    25           61

7    14    7    49    49           98

8    16    9    64    81         145

8 & 9

Answer 9

let x - one positive integer..

 y - other integer

x = 2y - 7 eq1

x^2 + y^2 = 145 eq2

plug in eq1 to eq2 

(2y - 7)^2 + y^2 = 145

4y^2 - 28y + 49 +y^2- 145 = 0

5y^2 - 28y - 96 = 0

 (5y + 12)(y - 8) = 0

y = - 12/5, y = 8

solving for x

when y = 8

x = 2y - 7

x = 2(8) - 7 = 9

when y = - 12/5

x = 2(-12/5) - 7 = -59/5

The possible positive integers are 9 and 8...//

Answer 10

Let x = first positive integer

Let y = second positive integer

The first is 7 less than twice the second:

x = 2y - 7

The sum of the squares is 145:

x² + y² = 145

We now have a system of two equations and two unknowns so we can solve with substitution:

(2y - 7)² + y² = 145

4y² - 28y + 49 + y² = 145

5y² - 28y + 49 = 145

5y² - 28y - 96 = 0

Quadratic equation:

y = [ -b ± √(b² - 4ac)] / (2a)

y = [ -(-28) ± √((-28)² - 4(5)(-96))] / (2 * 5)

y = [ 28 ± √(784 + 1920)] / 10

y = [ 28 ± √(2704)] / 10

y = (28 ± 52) / 10

y = -24/10 and 80/10

y = -12/5 and 8

Since y is said to be a positive integer we can throw out the negative fraction to leave us with:

y = 8

Now we can solve for x:

x = 2y - 7

x = 2(8) - 7

x = 16 - 7

x = 9

Your numbers are 9 and 8.