Animated Solution for Mathematics - Quadratic Equations: STATEMENT-1: $(p^2-q)(b^2-ac)\ge 0$ and STATEMENT-2: $b\ne pa$ or $c\ne qa$

We are given two quadratic equations: $E_1:a{x}^2+2bx+c=0$ $E_2:x^2+2px+q=0$

These equations share a common root, which we shall denote as $\alpha$. Geometrically, this implies that the two parabolas intersect the $x$-axis at the same point $\alpha$.

Our Compass

For any quadratic equation $A{x}^2+Bx+C=0$, the discriminant is defined as $D=B^2-4AC$. This value dictates the nature of the roots.

For $E_1$, the discriminant is:

For $E_2$, the discriminant is:

The product of these discriminants is:

If the common root $\alpha$ is a real number, then both parabolas must intersect or touch the $x$-axis. This condition forces both discriminants to be non-negative: $D_1\ge 0$ and $D_2\ge 0$

Since both values are non-negative, their product must also be non-negative:

This confirms that the inequality holds true when the roots are real.

If the roots are complex, they must appear in conjugate pairs because the coefficients are real. If the equations share one complex root, they must share the entire conjugate pair.

In this scenario, both $D_1<0$ and $D_2<0$. The product of two negative numbers is positive:

Thus, the inequality $(b^2-ac)(p^2-q)\ge 0$ remains valid. Whether the roots are real or complex, the inequality is an unbreakable law of this system.

Statement 2

Statement 2 asserts that $b eq pa$ or $c eq qa$. This condition ensures that the two equations are not identical.

If $b=pa$ and $c=qa$, the equations would be scalar multiples of each other, representing the same parabola. While this statement is mathematically true for distinct equations, it is not the underlying cause of the inequality in Statement 1.

Conclusion: Statement 1 is a direct consequence of the properties of discriminants and conjugate roots. Statement 2 provides a condition for distinctness but does not serve as the logical explanation for the validity of Statement 1.