Factoring: Number Problems Solving! UHH!!?

Question

Numba 2: When a number is added to six times its square, the result is 12. Find the NUmber

Numba 4: The sum of the squares of two consecutive integers is 41
Find the integers

Numba 5: The sum of the squares of three consecutive, posotive integers is equal to the sum of the squares of the next two integers
Find the five integers

Numba 6 Find two consecutive even integers whiose product is 80

numba 7 Twice the square of a certain positive number is 144 more than twice the number What is the Number?

Numba 8 The square of a positive number decreased by 10 is 2 more than 4 times the number What is the number??

Yea I know the spelling sucks but all the numbers are right!
Show all work
Set up and solve
Thanks!

Answer 1

2) x + 6x^2 = 12
6x^2 + x -12=0
x^2 + 1/6 x -2=0
x^2 + 1/6x + (1/12)^2 - (1/12)^2 - 2=0
(x+1/12)^2 - (1/144 + 288/144) = 0
(x+1/12)^2 - 289/144=0
(x+1/12 - 17/12)(x+1/12+17/12)=0
(x - 16/12)(x+ 18/12)=0
x= 16/12 = 4/3
x=-18/12= - 3/2

4) x^2 + y^2 = 41 where y = x+1
x^2 + x^2 + 2x + 1 = 41
2x^2 + 2x - 40 = 0
x^2 + x - 20 = 0
x^2 + x + (1/2)^2 - (1/2)^2 - 20 = 0
(x+1/2)^2 - (1/4 + 80/4) = 0
(x + 1/2)^2 - 81/4 = 0
(x + 1/2 - 9/2)(x + 1/2 + 9/2) = 0
(x - 4)(x + 5)=0
so if x = 4 then y=5
and if x= - 5 then y= -4

Answer 2

No.2 : x + 6x^2 = 12
Rearrange : 6x^2 + x - 12 = 0
Factorise : (2x + 3)(3x - 4) = 0
Set each term equal to zero :
2x + 3 = 0 or 3x - 4 = 0 implies x = -3/2 or 4/3.

No.4 : x^2 + (x + 1)^2 = 41
Expand : x^2 + x^2 + 2x + 1 = 41
2x^2 + 2x - 40 = 0
Divide through by 2 :
x^2 + x - 20 = 0
Factorise : (x - 4)(x + 5) = 0
Therefore, x = 4 or -5

No.5 : x^2 + (x + 1)^2 + (x + 2)^2 = (x + 3)^2 + (x + 4)^2
Expand :
x^2 + x^2 + 2x + 1 + x^2 + 4x + 4 = x^2 + 6x + 9 + x^2 + 8x + 16
Sum like terms : 3x^2 + 6x + 5 = 2x^2 + 14x + 25
Subtract RHS from LHS : x^2 - 8x - 20 = 0
Factorise : (x + 2)(x - 10) = 0
Therefore, x = -2 or 10
Integers required to be positive, so x = 10.
Thus : 10^2 + 11^2 + 12^2 = 13^2 + 14^2

No.6 : x * (x + 2) = 80
Multiply out and rearrange : x^2 + 2x - 80 = 0
Factorise : (x + 10)(x - 8) = 0
Therefore, x = -10 or 8.
Thus, (-10)(-8) = 80, or (8)(10) = 80.

No.7 : 2x^2 = 144 + 2x
Divide through by 2 : x^2 = 72 + x
Rearrange : x^2 - x - 72 = 0
Factorise : (x + 8)( x - 9) = 0
Therefore, x = -8 or 9, but x must be positive, so x = 9.

No.8 : x^2 - 10 = 2 + 4x
Subtract RHS from LHS : x^2 - 4x - 12 = 0
Factorise : (x + 2)(x - 6) = 0
Therefore, x = -2 or 6, but x > 0, so x = 6.

Answer 3

#2 (I might have this one wrong, or it didn't add up to a whole number, but nevertheless...)
x + 6*x^2 = 12
solving, x=4/3, or -3/2

#4
x^2 + y^2 = 41
the only possible solution via trial and error is 4, 5

#5 takes too much thinking for my brain now ;)

#6 8, 10 -> 8*10 = 80

#7
2x^2 = 144 + 2x
solving, x = 8, -9, the latter not being the one you want

#8
x^2 - 10 = 2 + 4x
solving, x = 6, -2, the latter not being the one you want

Answer 4

8x^2 - 34x + 24 = -11 8x^2 - 34x + 24 + 11 = 0 8x^2 - 34x + 35 = 0 (2x - 5)(4x - 7) = 0 2x - 5 = 0 2x = 5 x = 5/2 (2.5) 4x - 7 = 0 4x = 7 x = 7/4 (a million.seventy 5) ? x = 5/2 (2.5) , 7/4 (a million.seventy 5)

Answer 5

4^2+5^2=41

8x10 = 80

That's for now