Animated Solution for Mathematics - Quadratic Equations: Consider the two sets: $A=\{m\in R:\text{both the roots of }{x}^2-(m+1)x+m+4=0\text{ are real}\}$ and $B=[-3,5)$. Which of the following is not true?

To unlock the solution, we must first determine the condition for real roots of the quadratic equation $x^2-(m+1)x+m+4=0$. In JEE mathematics, the discriminant $D=b^2-4ac$ serves as the gatekeeper for the nature of roots.

For the roots to be real, we require the condition $D\ge 0$. Identifying the coefficients, we have $a=1$, $b=-(m+1)$, and $c=m+4$.

Substituting these into the discriminant formula, we obtain:

Expanding this expression yields:

Simplifying the terms, we arrive at the quadratic expression:

Visualizing the Inequality

We now solve the inequality $m^2-2m-15\ge 0$. Factoring the quadratic, we get:

The critical points are $m=-3$ and $m=5$. Applying the Wavy Curve Method, we test the intervals to determine where the expression is non-negative.

The expression is positive for $m\ge 5$ and $m\le -3$. Therefore, Set $A$ is defined as:

The use of square brackets is essential here, as the inequality includes the equality case where $D=0$.

We are given Set $B=[-3,5)$. This is a bounded interval starting at $-3$ (inclusive) and ending at $5$ (exclusive). We now perform the requested set operations.

Intersection ($A\cap B$): Since $A$ contains $-3$ and $B$ contains $-3$, they share this single point. The interval $B$ lies entirely between the two components of $A$, except for the shared boundary at $-3$. Thus, $A\cap B=\{-3\}$.

Difference ($B-A$): We take $B$ and remove any elements belonging to $A$. Since $-3$ is in $A$, we remove it from $B$. The remaining interval is $(-3,5)$.

Union ($A\cup B$): Combining these sets fills the gap between the two rays of $A$. The result is the entire real number line, $\mathbb{R}$.

Finally, we evaluate $A-B$ by taking $A=(-\infty ,-3]\cup [5,\infty )$ and removing elements of $B=[-3,5)$.

First, the point $-3$ is present in $B$, so we must remove it from $A$. This transforms the closed bracket at $-3$ into an open one, resulting in $(-\infty ,-3)$.

Next, we examine the point $5$. Since $B=[-3,5)$ excludes $5$, the value $5$ is not removed from $A$. Therefore, $5$ remains included in the final set.

The resulting set is:

A - B = (-\infty, -3) \cup