Animated Solution for Mathematics - Quadratic Equations: Let $a,b\in R,a\ne 0$, such that the equation, $a{x}^2-2bx+5=0$ has a repeated root $\alpha$, which is also a root of the equation $x^2-2bx-10=0$. If $\beta$ is the root of this equation, then ${\alpha }^2+{\beta }^2$ is equal to:

Welcome, future engineer. Today, we are not just solving an equation; we are uncovering a hidden symmetry. We have two quadratic equations, $a{x}^2-2bx+5=0$ and $x^2-2bx-10=0$.

They seem like strangers, but they share a secret: a common root, $\alpha$. This $\alpha$ is the bridge between these two mathematical worlds. Let us walk across it together.

Look at the first equation: $a{x}^2-2bx+5=0$. We are told it has a repeated root, $\alpha$. In the language of polynomials, this is a gift. It means the roots are $\alpha$ and $\alpha$.

Using Vieta's formulas, we know the sum of the roots is:

Now, consider the product of the roots. The product is $\alpha \cdot \alpha ={\alpha }^2$. From the equation, the product is also $\frac{5}{a}$.

So, we have established a vital relationship: ${\alpha }^2=\frac{5}{a}$. If we square our expression for $\alpha$, we get:

Keep this relationship safe; it is the key to unlocking the entire problem.

Now, we turn our attention to the second equation: $x^2-2bx-10=0$. We know $\alpha$ is a root here as well. This means if we plug $\alpha$ into this equation, it must balance perfectly:

This is where the magic happens. We have expressions for ${\alpha }^2$ and $\alpha$ from our first phase. Let us substitute them in.

Replacing ${\alpha }^2$ with $\frac{5}{a}$ and $\alpha$ with $\frac{b}{a}$, our equation transforms into:

Do you see it? We have a $b^2$ term, and we already know that $b^2=5a$. Let us substitute that into our equation:

The $a$ terms cancel out beautifully, leaving us with:

With $a$ in our hands, the rest of the puzzle falls into place. We know ${\alpha }^2=\frac{5}{a}$. Substituting $a=\frac{1}{4}$, we get ${\alpha }^2=20$.

We are almost at the finish line. We need ${\alpha }^2+{\beta }^2$. We have ${\alpha }^2=20$.

What about ${\beta }^2$? Look at the second equation again: $x^2-2bx-10=0$. The product of its roots is $\alpha \cdot \beta =-10$.

Squaring both sides, we get:

Since we know ${\alpha }^2=20$, we have $20\cdot {\beta }^2=100$, which gives us ${\beta }^2=5$. Finally, we add them together:

Take a moment to appreciate this. We started with two seemingly independent equations and, by following the trail of the common root, we dismantled the complexity layer by layer. This is the essence of JEE Advanced mathematics—not just calculation, but the art of connecting the dots. Well done.