Question 61

The minimum value of the sum of the squares of the roots of $$x^2 + (3-a)x = 2(a-1)$$ is

Given equation,

$$x^2+(3-a)x=2(a-1)$$

or,

$$x^2+(3-a)x-2(a-1)=0$$

Let the roots be

$$\alpha,\beta$$

Then,

$$\alpha+\beta=a-3$$

and

$$\alpha\beta=-2(a-1)$$

Now,

$$\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$$

$$=(a-3)^2-2[-2(a-1)]$$

$$=a^2-6a+9+4a-4$$

$$=a^2-2a+5$$

Complete the square:

$$a^2-2a+5=(a-1)^2+4$$

Since

$$(a-1)^2\ge0,$$

minimum value occurs at

$$a=1$$

Hence, minimum value is

$$\boxed{4}$$.

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