# The nature of the roots of a quadratic equation

• Jul 16th 2013, 08:00 AM
Seaniboy
The nature of the roots of a quadratic equation
Hi,
Thanks.

Show that the roots of the equation $x^2 - (a + d)x + (ad - b^2) = 0$ are real.
• Jul 16th 2013, 08:20 AM
Prove It
Re: The nature of the roots of a quadratic equation
Have you tried checking the discriminant of this quadratic equation?
• Jul 16th 2013, 12:42 PM
Seaniboy
Re: The nature of the roots of a quadratic equation
Yes Sir, I have. $b^2 - 4ac$ greater than or equal to zero;

$a = 1, b = a+d, c = ad - b^2$;

$b^2 = (a + d)^2 = a^2 + 2ad + d^2$;

$- 4ac = - 4(1)(ad - b^2) = - 4ad + 4b^2$;

However, how does one prove that $a^2 - 2ad + d^2 + 4b^2$ is greater than or equal to zero?
• Jul 16th 2013, 12:47 PM
MarkFL
Re: The nature of the roots of a quadratic equation
Try factoring the first 3 terms as the square of a binomial...then you will have the sum of two squares...
• Jul 16th 2013, 01:10 PM
Plato
Re: The nature of the roots of a quadratic equation
Quote:

Originally Posted by Seaniboy
Show that the roots of the equation $x^2 - (a + d)x + (ad - b^2) = 0$ are real.

You are making it too hard.
$\\(a+d)^2-4(1)(ad-b^2)\\=a^2+2ad+d^2-4ad+4b^2\\=a^2-2ad+d^2+4b^2\\=(a-d)^2+4b^2$
• Jul 17th 2013, 05:52 AM
Seaniboy
Re: The nature of the roots of a quadratic equation
Thanks Plato.
• Jul 17th 2013, 05:53 AM
Seaniboy
Re: The nature of the roots of a quadratic equation
Thanks Mark.