Animated Solution for Mathematics - Quadratic Equations: If $\alpha \ne \beta$ but ${\alpha }^2=5\alpha -3$ and ${\beta }^2=5\beta -3$ then the equation having $\alpha /\beta$ and $\beta /\alpha$ as its roots is

Welcome, future engineer. Today, we are going to peel back the layers of a classic quadratic problem. It is not just about finding an answer; it is about recognizing the hidden architecture of algebra.

Imagine you are staring at the equations ${\alpha }^2=5\alpha -3$ and ${\beta }^2=5\beta -3$. If we rearrange these, we get ${\alpha }^2-5\alpha +3=0$ and ${\beta }^2-5\beta +3=0$.

Both $\alpha$ and $\beta$ satisfy the condition $x^2-5x+3=0$. Since we are told $\alpha eq \beta$, we have identified the parent equation for both roots.

We do not need to solve for $\alpha$ or $\beta$ individually, as that would be a tedious path. Instead, we use the power of Vieta's formulas for the quadratic equation $x^2-5x+3=0$.

The sum of the roots is:

The product of the roots is:

We shift our focus to the target: a new quadratic equation whose roots are $r_1=\frac{\alpha }{\beta }$ and $r_2=\frac{\beta }{\alpha }$. To build any quadratic equation, we use the general form $x^2-Sx+P=0$, where $S$ is the sum and $P$ is the product of the roots.

First, we calculate the sum $S=\frac{\alpha }{\beta }+\frac{\beta }{\alpha }$. Finding a common denominator, we get:

Using the identity ${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta$, we substitute our known values:

Next, we calculate the product $P=\frac{\alpha }{\beta }\cdot \frac{\beta }{\alpha }$. Through cancellation, we find:

With $S=\frac{19}{3}$ and $P=1$, our equation is $x^2-\frac{19}{3}x+1=0$. Multiplying the entire equation by $3$ to clear the fraction, we obtain the final result: