Question

If a,b are real numbers such that ab-4a+4b=18, find the minimum value of

a^2+b^2

Collected in the board: Quadratic function

Steven Zheng posted 12 hours ago

Answer

Let

y = a^2+b^2
(1)
x = a-b
(2)

x^2-y = (a-b)^2 - (a^2+b^2) = -2ab

Then

ab = \dfrac{1}{2}(y-x^2)
(3)

Onthe other hand,

ab-4a+4b=ab - 4x = 18

ab = 18+4x
(4)

A quadratic equation is obtained from (3) and (4)

\dfrac{1}{2}(y-x^2) = 18+4x

y = x^2+8x+36 = (x+4)^2+20\geq 20

which shows y_{min} = 20 when x = -4

From (2) and (4), the following system of equations is obtained

\begin{cases} a-b &= -4 \\ ab &= 2 \end{cases}

Solving the system of equations gives

a = -2+\sqrt{6}, b = 2+\sqrt{6} or a = -2-\sqrt{6}, b = 2-\sqrt{6}

Therefore,

The minimum value of a^2+b^2 is 20


Steven Zheng posted 12 hours ago

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