# How do you turn any quadratic into squared form [duplicate]

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.

## marked as duplicate by user296602, N. F. Taussig, Namaste algebra-precalculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Dec 4 '18 at 21:45

• The magic words are "completing the square". en.wikipedia.org/wiki/Completing_the_square – Arturo Magidin Dec 4 '18 at 21:02
• Yeah, I didn't see the 0.75 so I was confused but it's vertex form isn't it lol – ming Dec 4 '18 at 21:03
• See also many questions tagged completing-the-square – user296602 Dec 4 '18 at 21:05
• $x^2$ 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. – Tito Eliatron Dec 4 '18 at 21:05
• Now that you realize this is the vertex form, do you still have a question? – David K Dec 4 '18 at 21:08

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