Verified answer

1.

Let

x - 1st number

x + 2 2nd number

x^2 + (x + 2)^2 = 20

x^2 + x^2 + 4x + 4 = 20

2x^2 + 4x - 16 = 0

x^2 + 2x - 8 = 0

(x + 4)(x - 2) = 0

So 1st no = -4, and 2

2nd no is -2 or 4

(1st, 2nd) = (-4, -2), (2, 4)

2.

Let x = 1st no.

8 - x = 2nd no

x^2 + (8 - x)^2 = 170

x^2 + 64 - 16x + x^2 = 170

2x^2 - 16x - 106 = 0

x^2 - 8x - 53 = 0

No solution.

You might have mistyped the sum as 8 when it should be 18.

3.

Let x = 1st no.

x + 4 = 2nd no.

(x + 4)^2 - x^2 = 88

x^2 + 8x + 16 -x^2 = 88

8x = 72

x = 9

x + 4 = 13

4.

Let x = one number

x + 3 = other number

x(x + 3) + 49 = 89

x^2 + 3x + 49 = 89

x^2 + 3x - 40 = 0

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

x + 8 = 0

x = - 8 one number

-8 + 3 = -5 other number

(x - 5) = 0

x = 5 one number

5 + 3 = 8 other number

Two sets of answers (-8, -5) and (5, 8)

DOING WITHOUT DOING (take it informally please)

1. 20 = 4^2 + 2^2 and 4 -- 2 = 2. numbers are 4 and 2.

2. 170 = 12^2 + 6^2 and 12 + 6 = 18. numbers are 12 and 6.

3. 88 = 13^2 -- 9^2 and 13 -- 9 = 4. numbers are 13 and 9.

4. 89 = 49 + 8*5 and 8 = 5 + 3. numbers are 8 and 5.

1. x-y = 2

x^2 + y^2 = 20

since x-y = 2, x = y+2

,', (y+2)^2 + y^2 = 20

2y^2+ 2y + 4 = 20

2y^2 + 2y - 16 = 0

2. x + y = 18

which implies x = 18 - y

(18-y)^2 + y^2 = 170

2y^2 - 36y + 154 = 0

OR

y^2 - 18y + 77 = 0

3. x - y = 4

which implies x = y+4

8y + 16 = 88

8y = 72

y = 9

,', x = 13

4. x = 3y

49 + xy = 89

,', 49 + 3y^2 = 89

,', y = root of (40/3) [?!?!]

[obviously, this doesn't look right. sorry!]

1.

working equations:

"The difference of two numbers is 2"

A-B = 2 [eq 1]

"the sum of their squares is 20"

A^2 + B^2 = 20 [eq 2]

.

1... one of them is 2 and the other is 4...

2^2 = 4 .... 4^2 = 16...

2... not sure... might have to use -ve numbers???

3... 13 and 9.

4...not quite sure