Animated Solution for Mathematics - Quadratic Equations: If $x^2+(a-b)x+(1-a-b)=0$ where $a,b\in R$ then find the values of $a$ for which equation has unequal real roots for all values of $b$.

We are given the quadratic equation in $x$: We are tasked with finding the values of $a$ such that this equation possesses unequal real roots for all real values of $b$.

For a quadratic equation to have unequal real roots, its discriminant $D_x$ must be strictly greater than zero. Given the standard form $A{x}^2+Bx+C=0$, where $A=1$, $B=(a-b)$, and $C=(1-a-b)$, the discriminant is defined as $D_x=B^2-4AC$.

Substituting our coefficients, we obtain:

Expanding the expression for $D_x$, we get:

Since the condition must hold for all $b\in \mathbb{R}$, we treat $D_x$ as a quadratic function of $b$. Rearranging the terms, we have:

We require $D_x>0$ for all $b$. Geometrically, this represents a parabola opening upwards (since the coefficient of $b^2$ is $1>0$) that must lie entirely above the horizontal axis.

For this to occur, the quadratic in $b$ must have no real roots. Consequently, the discriminant of this new quadratic, denoted as $D_b$, must be strictly less than zero.

We calculate $D_b$ using the coefficients of our $b$-quadratic: $A=1$, $B=(4-2a)$, and $C=(a^2+4a-4)$.

Expanding the terms:

Summing these components, the $a^2$ terms cancel out:

Applying the condition $D_b<0$:

The condition for the equation to have unequal real roots for all $b$ is $a>1$.