Let a,b be 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 5 months ago



x = a-b
y = a^2+b^2

Taking square of equation (1) gives

x^2 =a^2+b^2-2ab

Substitute (2) and solve for ab

ab = \dfrac{y-x^2}{2}

Substitute (4) and(1) to given equation ab-4a+4b=18, then

\dfrac{y-x^2}{2}-4x = 18

Rearrange terms to the form of a quadratic function.

\begin{aligned} y&= x^2+8x+36 \\ &= (x+4)^2+20 \end{aligned}

Therefore, the minimum value of y, that is, a^2+b^2 is 20 when x= -4

Steven Zheng posted 5 months ago

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