If $x^2 + 4x = 10$, evaluate the expression $E = (x + 3)^2 + (x+1)^2$

I don't know what these type of problems are called, hence the odd title and tags. I'd like somebody to tell me what sort of problem this is (what it's called) so I can do self-research or give me some sort of hint to point me in the direction of solving it.

  • $\begingroup$ I would just call this "polynomial algebra", or something like that. In this case, you can expand to get $E[x]=x^2+6x+9+x^2+2x+1=2x^2+8x+10$ and at this point you can use the equation that $x$ is known to satisy. $\endgroup$
    – lulu
    Sep 8, 2015 at 23:30
  • $\begingroup$ I see, but why is there a "[x]" in front of the E in "E[x]=x2+6x+9+x2+2x+1=2x2+8x+10?" $\endgroup$
    – Ben
    Sep 8, 2015 at 23:32
  • $\begingroup$ In my expression? I just meant to indicate that $E$ was a function of $x$. Probably should have omitted it. $\endgroup$
    – lulu
    Sep 8, 2015 at 23:33

1 Answer 1


If the powers used are less or equal to two, you should search for quadratic equations / quadratic functions. You should also know, if you don't already, the Formulas of abridged multiplication.

For your particular question, you only need the $(a+b)^2=a^2+2ab+b^2$ formula and the basic grouping rules. By using it, you get:

E=$x^2+6x+9+x^2+2x+1=2x^2+8x+10=2(x^2+4x)+10=2 \cdot 10 +10=30$

  • $\begingroup$ I understand what you did up to the point where you transition from 2x^2+8x+10 to 2(x^2+4)+10 = 2⋅10+10=30 $\endgroup$
    – Ben
    Sep 8, 2015 at 23:47
  • $\begingroup$ It was just a typo! Fixed now! (First equality is due to factoring a 2, second is replacing the bracket with the value given) $\endgroup$
    – H3llShadow
    Sep 8, 2015 at 23:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .