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

We are given the quadratic equation $x^2-2ax+a^2+a-3=0$. We are tasked with finding the range of $a$ such that both roots are real and strictly less than $3$.

Let the function be defined as $f(x)=x^2-2ax+a^2+a-3$. Since the leading coefficient is $1>0$, the graph is a parabola opening upwards.

To ensure both roots $\alpha$ and $\beta$ are real and satisfy $\alpha ,\beta <3$, we must satisfy three specific conditions simultaneously:

1. Existence of Roots: The discriminant $D$ must be non-negative to ensure the roots are real. Thus, $D\ge 0$.

2. Symmetry Constraint: The axis of symmetry (the x-coordinate of the vertex) must lie to the left of the boundary $x=3$. Thus, $x_v<3$.

3. Boundary Condition: Since the parabola opens upwards, the value of the function at the boundary $x=3$ must be positive to ensure the roots do not cross into the region $x\ge 3$. Thus, $f(3)>0$.

First, we calculate the discriminant $D=b^2-4ac$:

Setting $D\ge 0$ gives $-4a+12\ge 0$, which simplifies to $a\le 3$.

Next, we determine the vertex position $x_v=\frac{-b}{2a}$:

Applying the condition $x_v<3$, we obtain $a<3$.

Finally, we evaluate the boundary condition $f(3)>0$:

Setting $a^2-5a+6>0$ and factoring, we get $(a-2)(a-3)>0$. This inequality holds when $a<2$ or $a>3$.

To find the valid range for $a$, we take the intersection of our three constraints:

1. $a\le 3$ 2. $a<3$ 3. $a<2$ or $a>3$

The intersection of these conditions is the region where all three overlap. Therefore, the final solution is: