Wednesday

October 22, 2014

October 22, 2014

Posted by **kim** on Sunday, December 3, 2006 at 6:13pm.

Prove that the roots of

ax^2 + (a + b)x+b are real for all values of k

note the "x"s aren't multiplication signs.

a x^2 + bx + c has the discriminant of

D = b^2 - 4ac.

If D is nonnegative then the function has real roots.

In this case you have

D = (a + b)^2 - 4 a b = (a-b)^2

which is larger than or equal to zero because it is a square.

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