Linear problem

[Guest]

Determine consecutive integers whose difference of cubes is 91.

[Baltuilhe]

Good morning!

[tex](x+1)^3-x^3=91\\
x^3+3x^2+3x+1-x^3=91\\
3x^2+3x+1=91\\
3x^2+3x-90=0\;\div 3\\
x^3+x-30=0\\
\Delta=(1)^2-4(1)(-30)=1+120=121\\
x=\frac{-(1)\pm\sqrt{121}}{2(1)}\\
x=\frac{-1\pm 11}{2}\\
x'=\frac{-1-11}{2}=-6\\
x''=\frac{-1+11}{2}=5[/tex]

So, 5 and 6!

[HallsofIvy]

My only question is "Why are you calling this a linear problem?"


Where you have [tex]x^2+ x- 30= 0[/tex] it simpler to factor [tex](x+ 6)(x- 5)= 0[/tex]
and get x= -6 and x= 5.

You use the quadratic formula to get those roots but then erroneously write
"So, 5 and 6!"

"

[Baltuilhe]

My only question is "Why are you calling this a linear problem?"


Where you have [tex]x^2+ x- 30= 0[/tex] it simpler to factor [tex](x+ 6)(x- 5)= 0[/tex]
and get x= -6 and x= 5.

You use the quadratic formula to get those roots but then erroneously write
"So, 5 and 6!"

"

The question was: "Determine consecutive integers whose difference of cubes is 91"
So, 5 and 6 are consecutive integers that reach the answer.
But, as integers can be negative, too, -6 and -5 are a valid pair of answers, too. Sorry for that