Thursday

July 31, 2014

July 31, 2014

Posted by **Shay** on Thursday, January 18, 2007 at 5:27pm.

x^2+(k-2)x-2k=0

has equal and real roots.

a x^2 + bx + c = 0 has two different roots if the discriminant D defined as:

D = b^2 - 4 a c does not equal zero.

If D = 0 then there is one root. That root is then real if b/a is real. In this case this means that:

(k-2)^2 + 8k = 0 --->

(k+2)^2 = 0 --->

k = -2

k must also be real and -2 is real, so k = -2 is a valid solution.

- algebra -
**Anonymous**, Thursday, October 4, 2007 at 9:05pmits -3

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