Advanced High-School Mathematics

SECTION 3.5 Quadratic Discriminant 163

Solving for the discriminant in terms ofm, we quickly obtain

0 = b^2 − 4 ac

= (−12)^2 − 4 · 2 ·(23−m)

= 144−8(23−m)

= 144−184 + 8m=−40 + 8m

and so one obtains the minimum value ofm= 5.Of course, this is

not the “usual” way students are taught to find the extreme values

of a quadratic function: they use the method of “completing the

square” (another useful technique).

Example 2. Here’s an ostensibly harder problem. Find the minimum

value of the functiong(x) =x+

1

x

, x >0. Before going further,

note that the ideas of Section 3.1 apply very naturally: from

0 ≤

Ñ

√

x−

1

√

x

é 2

=x+

1

x

− 2

we see immediately thatx+

1

x

≥ 2 with equality precisely when

x= 1. That is the say, the minimum value of the objective function

x+

1

x

is 2.

Solution. Denoting this minimum by

m, then the graph of y = m will

again be tangent to the graph of

y = g(x) where this minimum oc-

curs. Here, the equation isx+

1

x

=

m, which quickly transforms to the

quadratic equationx^2 −mx+ 1 = 0.

For tangency to occur, one must

have that the discriminant of

x^2 −mx+ 1 vanishes.

y 6

x

y =x+

1

x

D < 0

D = 0

D > 0