If the roots of the equation x^2 - 2ax + a^2 | vedclass

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Question

(A) Let $f(x) = x^2 - 2ax + a^2 + a - 3$.For the roots to be real and less than $3$,we must satisfy three conditions: $1$. Discriminant $D \geq 0$: $D

Vedclass · Accepted Answer

$a < 2$

Vedclass · Answer

(A) Let $f(x) = x^2 - 2ax + a^2 + a - 3$.For the roots to be real and less than $3$,we must satisfy three conditions: $1$. Discriminant $D \geq 0$: $D = (-2a)^2 - 4(1)(a^2 + a - 3) = 4a^2 - 4a^2 - 4a + 12 = 12 - 4a \geq 0 \implies a \leq 3$.$2$. Vertex $x_v $x_v = -b/(2a) = 2a/2 = a $3$. $f(3) > 0$: $f(3) = 3^2 - 2a(3) + a^2 + a - 3 = 9 - 6a + a^2 + a - 3 = a^2 - 5a + 6 > 0$.$(a - 2)(a - 3) > 0 \implies a  3$.Combining all conditions: $a \leq 3$ $AND$ $a  3$).The intersection is $a < 2$.

If the roots of the equation $x^2 - 2ax + a^2 + a - 3 = 0$ are real and less than $3$,then:

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