Animated Solution for Mathematics - Quadratic Equations: The number of integral values of m for which the equation $(1+m^2)x^2-2(1+3m)x+(1+8m)=0$ has no real root is :

Understanding 'No Real Roots'

Welcome, future engineer. Today, we are going to dissect a problem that seems like a simple algebraic exercise but is actually a beautiful study of the behavior of quadratic functions.

We are given the equation $(1+m^2)x^2-2(1+3m)x+(1+8m)=0$ and asked to find the number of integral values of $m$ for which it has no real roots.

Imagine a parabola on a coordinate plane. If an equation has 'no real roots,' it means the parabola is floating—it never touches or crosses the x-axis.

Mathematically, this 'floating' state is governed by the discriminant, $D=B^2-4AC$. If $D<0$, the parabola stays strictly above or below the axis, and the roots remain in the complex plane.

Let us identify our players. Comparing our equation to the standard form $A{x}^2+Bx+C=0$, we find: $A=(1+m^2)$ $B=-2(1+3m)$ $C=(1+8m)$

Now, we invoke the condition for no real roots: $D<0$. Substituting our coefficients, we get the inequality:

I know, it looks intimidating. But take a deep breath; in JEE Advanced, the complexity is often a mask for a simple, elegant simplification waiting to be discovered.

Notice the factor of 4. We have a 4 coming from the square of the $-2$, and a 4 in the second term. Let us divide the entire inequality by 4 immediately.

We are left with:

Now, expand with precision. The first term becomes $1+6m+9m^2$. The second term, when multiplied out, yields $1+8m+m^2+8m^3$.

Subtracting these, the constants cancel out beautifully, leaving us with:

We are almost there. To make this easier to read, let us multiply by $-1$. Remember the golden rule: when you multiply an inequality by a negative number, the sign flips!

So, we obtain:

Now, factor out the common $2m$:

Look closely at that quadratic inside the parenthesis. It is a perfect square: $(2m-1{)}^2$. Thus, our inequality simplifies to:

An Infinite Possibility

We need this product to be positive. Since $(2m-1{)}^2$ is a square, it is always non-negative. It is zero only when $m=\frac{1}{2}$.

Since we need the product to be strictly greater than zero, we must exclude $m=\frac{1}{2}$. For all other values of $m$, $(2m-1{)}^2$ is positive.

Therefore, the inequality reduces to $2m>0$, which means $m>0$. We are looking for integral values of $m$ such that $m>0$ and $m eq \frac{1}{2}$.

The integers satisfying this are $1,2,3,4,\dots$ and so on. There is no upper bound! Thus, there are infinitely many integral values of $m$.