• Feb 22nd 2013, 04:22 AM
heaviside
I was doing some problems last night, and I came across this one and I am not sure how to get started.

If the polynomial \$\displaystyle Eq. A: ax^2+bx+c\$ does not have any (real*) roots, prove that \$\displaystyle Eq. B: ax^2+bx-c\$ has two (real*) roots.

*In full disclosure, the text did not include the word 'real' before the word 'roots'. Obviously, by the Fundamental Theorem of Algebra, both will have two roots.

I was thinking about arguing that equation A must have a discriminant that is negative (\$\displaystyle b^2-4ac < 0 \$) and therefore Equation B will have a positive discriminant (\$\displaystyle b^2-4a(-c) = b^2+4ac > 0 \$). A quadratic with a positive discriminant has two real roots.

Is there a better / more clever way to solve this?

Thanks
• Feb 22nd 2013, 06:01 AM
HallsofIvy
Okay, but you are skipping over the crucial part of the proof: how does \$\displaystyle b^2- 4a(-c)> 0\$ follow from \$\displaystyle b^2- 4ac< 0\$?
• Feb 22nd 2013, 06:52 AM
MINOANMAN
since the quadratic ax^2+bx+c=0 has no real roots b^2<4ac .and since b^2 is positive either both a and c are positive or both are negative.In the second quadratic the c is negative therefore the product 4ac becomes now negative therefore b^2<4a(-c) is not possible since b^2 is a positive number .Therefore b^2-4a(-c)>0 and in this case the quadratic has two real roots.
• Feb 22nd 2013, 07:14 AM
heaviside