Animated Solution for Mathematics - Quadratic Equations: The integer '$k$', for which the inequality $x^2-2(3k-1)x+8k^2-7>0$ is valid for every $x$ in $R$ is :

Imagine you are standing in a vast, empty coordinate plane. You are given a quadratic expression:

The problem asks us to find an integer $k$ such that this expression remains strictly greater than zero for every possible real value of $x$. Since the coefficient of $x^2$ is $1$ (which is positive), we are dealing with an upward-opening parabola.

If this parabola were to touch or cross the $x$-axis, there would be values of $x$ where $f(x)$ is zero or negative. To ensure $f(x)>0$ for all $x$, the parabola must be a 'floating bowl'—hovering entirely above the $x$-axis without making contact.

The Discriminant

How do we mathematically enforce this 'floating' behavior? This is where the discriminant, $\Delta =b^2-4ac$, becomes our most powerful tool.

The discriminant tells us about the nature of the roots. If $\Delta >0$, the parabola crosses the axis twice; if $\Delta =0$, it touches the axis at one point.

However, if $\Delta <0$, the parabola has no real roots, meaning it never touches the $x$-axis. This is exactly the condition we require.

Let us perform the calculation. Our coefficients are $a=1$, $b=-2(3k-1)$, and $c=8k^2-7$. Substituting these into the discriminant formula:

Expanding this, we get:

We can simplify this immediately by dividing the entire inequality by $4$, which leaves us with:

Now, let us expand the square:

Grouping the terms, we find $9k^2-8k^2-6k+1+7<0$, which simplifies beautifully to:

We are left with a simple quadratic inequality: $k^2-6k+8<0$. Factoring this, we get:

Using the Wavy Curve Method, we identify the roots at $k=2$ and $k=4$. Since the parabola $g(k)=k^2-6k+8$ opens upwards, it is negative between its roots.

Thus, the valid range for $k$ is the open interval $(2,4)$. The question asks for the integer value of $k$ within this range.

Looking at the number line, the only integer strictly between $2$ and $4$ is $3$. By understanding the geometric soul of the quadratic, we have navigated the algebra and arrived at the solution: