Let \(p\) and \(q\) be roots of \(x^2 + px + q = 0\)are real. Prove that the roots of \(x^2 + px + q + (x + a)(2x + p) = 0\) are real, for any real number \(a\).

Guest May 9, 2020

#1**-1 **

Sicne the roots of x^2 +px + q are real, and the roots of (x + a)(2x + p) are real, the roots of x^2 + px + q + (x + a)(2x + p) are also real.

Guest May 9, 2020

#3**+1 **

Your question is not complete.

I can guess at the bits that are missing but I shouldn't have to do that.

Please check it properly.

Melody May 10, 2020

#4**+1 **

im sorry.

Let \(p\) and \(q\) be real numbers such that the roots of \(x^2 + px + q = 0\)are real. Prove that the roots of \(x^2 + px + q + (x + a)(2x + p) = 0\) are real, for any real number \(a\).

Guest May 10, 2020

#5**+1 **

Have you worked out what is wrong yet? I will leave you with that.

ALSO

You had no hope of following guests logic as he did not have any. He had no idea what he was talking about.

Here is my take:

Consider y=(x+a)(2x+p)

The roots for this are -a and -p/2 So (-p/2, 0) is a point on this parabola

Now consider.

\(y=x^2+px+q\\ \text{The axis of symmetry for this is }\frac{-p}{2}\\ \text{since this is a concave up parabola and the roots are real}\\ \text{It follows that when } x=\frac{-p}{2} \text{ the y value is negative or 0.}\\\)

When I add these two parabolas together the resulting y value when x=-p/2 MUST be equal to or less than 0.

The resultant parabola is concave up since the x^2 coefficient is positive (+3 to be precise).

Hence it must cross the x in 1 or 2 places and hence have 2 real roots (they can be equal roots)

Here is a graph that may help you understand.

https://www.desmos.com/calculator/g5ayxblz9f.

Byt the way, this is not a very good proof, it is more of a 'show'

Melody May 10, 2020