0
$\begingroup$

For example, $x^2 + 5x + 7$ is $(x + 2.5)^2 + 0.75$ but how would you figure that out? It's useful for proving any quadratic is greater than 0 but it's not always easy to find so. Thanks!

edit: Sorry I'm dumb I didn't see the + 0.75, this is just the vertex form.

$\endgroup$

marked as duplicate by user296602, N. F. Taussig, Namaste algebra-precalculus Dec 4 '18 at 21:45

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 2
    $\begingroup$ The magic words are "completing the square". en.wikipedia.org/wiki/Completing_the_square $\endgroup$ – Arturo Magidin Dec 4 '18 at 21:02
  • $\begingroup$ Yeah, I didn't see the 0.75 so I was confused but it's vertex form isn't it lol $\endgroup$ – ming Dec 4 '18 at 21:03
  • $\begingroup$ See also many questions tagged completing-the-square $\endgroup$ – user296602 Dec 4 '18 at 21:05
  • $\begingroup$ $x^2$ suggest you that teh fist term should be $x$. $5x$ suggest you that this shoud be the $2ab$ term, so it must be $(x+5/2)^2$. Finally, adjust the constant to get the same value. $\endgroup$ – Tito Eliatron Dec 4 '18 at 21:05
  • 1
    $\begingroup$ Now that you realize this is the vertex form, do you still have a question? $\endgroup$ – David K Dec 4 '18 at 21:08
0
$\begingroup$

Yes by completing the square we have that

$$x^2 + 5x + 7=x^2 + 5x + \frac{25}4-\frac{25}4+7=\left(x+\frac52\right)^2+\frac34\ge 0$$

since the two quantities are positive.

The same manipulation is also used for example to obtain the quadratic equation resolution formula and also in the context of quadratic forms.

$\endgroup$
  • $\begingroup$ Why the downvote? The expression before the editing in the OP was wrong. $\endgroup$ – gimusi Dec 4 '18 at 21:04

Not the answer you're looking for? Browse other questions tagged or ask your own question.