If the roots of the equation x2 – 2ax + a2 + a – 3 = 0 are real and less than 3, then
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a < 2
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2 ≤ a ≤ 3
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3 < a ≤ 4
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a > 4
Roots of x^2 - 2ax + a^2 + a - 3 = 0. D = 4a^2 - 4(a^2 + a - 3) = 4a^2 - 4a^2 - 4a + 12 = 12 - 4a. For real roots, D >= 0 => a <= 3. Roots are x = (2a +/- sqrt(12-4a))/2 = a +/- sqrt(3-a). Both roots < 3: a + sqrt(3-a) < 3 => sqrt(3-a) < 3-a. Let u = sqrt(3-a), u < u^2 => u^2 - u > 0 => u(u-1) > 0. u > 1 or u < 0. Since u >= 0, u > 1 => sqrt(3-a) > 1 => 3-a > 1 => a < 2.
For the equation x squared minus 2ax plus a squared plus a minus 3 equals 0, the sum of the roots is 2a. Because both roots must be less than 3, their sum must be less than 6, which gives 2a < 6 and a < 3. For the roots to be real, the discriminant must be greater than or equal to 0, so calculating b squared minus 4ac gives negative 4a squared minus 4 times 1 times a squared plus a minus 3. This simplifies to negative 4a plus 12, which must be greater than or equal to 0, meaning a is less than or equal to 3. Since any value of a less than 2 guarantees the sum 2a is strictly less than 4, it fully ensures both individual roots are less than 3.