Posted by Help generation on Saturday, September 26, 2009 at 3:39pm.
Express x2+4x+7 in the form (x+p)2+q. Hence show that equation x2+4x+7=0 has no real root???...I have problem in writing square in my computer so remember x2 is x square and 2 after bracket is squarer too.

math (Algebra)  Reiny, Saturday, September 26, 2009 at 5:55pm
the usual way to write powers in this forum is x^2 for x^{2}
so x^2 + 4x + 7
= x^2 + 4x + 4  4 + 7
= (x+2)^2 + 3
If I had the parabola y = (x+2)^2 + 3
its vertex would be above the xaxis at (2,3) and it opens upwards.
So it cannot cross the xaxis.
Since roots are simply the xintercepts of the corresponding function, we conclude that there are no real roots.
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