Given 2 consecutive integers, if 3 times the 1st is increased by the square of the second, the result is 85. Find these numbers

Answer 1

Two consecutive integers would be: 1st integer: x and 2nd integer: x+1, so with this knowledge we can then formulate the equation.

1. 3x+(x+1)²=85 this is is a quadratic so we will need to treat it as such when solving.
2. 3x+(x+1)(x+1)=85
3. 3x+x²+1x+1x+1=85 FOIL
4. x²+5x+1=85 combine like terms
5. x²+5x+1-85=85-85 subtract 85 from both sides to get into standard form Ax²+Bx+C=0
6. x²+5x-84=0 combine like terms
7. factors of -84:-1*84, -2*42, -3*28, -4*21, -6*14, -7*12
8. which two factor sets subtract to get +5? -7*12 so 12-7 =5 check.
9. (x-7)(x+12)=0 now we set each parenthesis = to 0 and solve to get two solutions.
10. x-7=0 so x=7
11. x+12=0 so x=-12
12. now plug in each number back into the original equation to see which ones work. 3x+(x+1)²=85
13. 3(7)+(7+1)²=85 gives 21+8² = 21+64 =85
14. 3(-12)+((-12)+1)²=85 gives -36+(-11)² = -36+121 =85
15. so since both solutions check out we can have x =7 or x =-12
16. the first consecutive set would be 7 and 7+1 =8 solution 1: 7,8
17. the second consecutive set would be -12 and -12+1=-11 solution 2: -12,-11