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 x-axis at (-2,3) and it opens upwards.

So it cannot cross the x-axis.

Since roots are simply the x-intercepts of the corresponding function, we conclude that there are no real roots.

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