# 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 \$\displaystyle 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. \$\displaystyle b^2 - 4ac\$ greater than or equal to zero;

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

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

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

However, how does one prove that \$\displaystyle 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 \$\displaystyle x^2 - (a + d)x + (ad - b^2) = 0\$ are real.

You are making it too hard.