posted by Help generation .
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.
the usual way to write powers in this forum is x^2 for x2
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.