The roots of the equation (q - r)x 2+ (r - p)x + (p - q) = 0 are
(p - q)/(q - r), 1
Finding the Roots of a Quadratic Equation
We are asked to find the roots of the given quadratic equation:
$(q - r)x^2 + (r - p)x + (p - q) = 0$
This equation is in the standard form of a quadratic equation, $ax^2 + bx + c = 0$, where:
- $a = q - r$
- $b = r - p$
- $c = p - q$
Analyzing the Coefficients
Let's look at the coefficients $a$, $b$, and $c$. A useful property for quadratic equations is to check the sum of the coefficients:
$a + b + c = (q - r) + (r - p) + (p - q)$
Let's simplify the sum:
$a + b + c = q - r + r - p + p - q$
We can see that the terms cancel out:
$a + b + c = (q - q) + (-r + r) + (-p + p) = 0 + 0 + 0 = 0$
So, the sum of the coefficients is zero.
Using the Property of Roots
For a quadratic equation $ax^2 + bx + c = 0$, if the sum of the coefficients ($a + b + c$) is equal to zero, then one of the roots of the equation is always 1.
Since $a + b + c = 0$ for the given equation, $x = 1$ is one root.
Finding the Second Root
For a quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$, we know the following relationships:
- Sum of roots: $\alpha + \beta = -b/a$
- Product of roots: $\alpha \times \beta = c/a$
We already found one root, say $\alpha = 1$. We can use the product of roots formula to find the other root, $\beta$.
$1 \times \beta = \frac{c}{a}$
$\beta = \frac{c}{a}$
Substitute the values of $a$ and $c$ from our equation:
$\beta = \frac{p - q}{q - r}$
So, the second root is $\frac{p - q}{q - r}$.
The Roots
The roots of the equation $(q - r)x^2 + (r - p)x + (p - q) = 0$ are $1$ and $\frac{p - q}{q - r}$.
Comparing these roots with the given options, we find that Option 2 lists the roots as $\frac{p - q}{q - r}$ and $1$.
| Coefficient | Value |
|---|---|
| $a$ | $q - r$ |
| $b$ | $r - p$ |
| $c$ | $p - q$ |
| Sum of Coefficients ($a+b+c$) | $(q-r)+(r-p)+(p-q) = 0$ |
| First Root | $1$ (since $a+b+c=0$) |
| Second Root ($c/a$) | $\frac{p-q}{q-r}$ |
Revision Table: Quadratic Equation Roots
| Property | Formula | Notes |
|---|---|---|
| Standard Form | $ax^2 + bx + c = 0$ | $a \neq 0$ |
| Sum of Roots ($\alpha + \beta$) | $-b/a$ | |
| Product of Roots ($\alpha \beta$) | $c/a$ | |
| Roots using Quadratic Formula | $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ | Discriminant $\Delta = b^2 - 4ac$ |
| Condition for $x=1$ as a root | $a + b + c = 0$ | Quick check |
Additional Information: Solving Quadratic Equations
There are several methods to solve quadratic equations, depending on the specific equation:
- Factoring: If the quadratic expression $ax^2 + bx + c$ can be factored into the form $(px + q)(rx + s)$, the roots are found by setting each factor to zero. This is often the fastest method when applicable.
- Completing the Square: This method involves rewriting the equation in the form $(x - h)^2 = k$ and then taking the square root of both sides. It's a foundational method used to derive the quadratic formula.
- Quadratic Formula: The most general method, the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, always provides the roots for any quadratic equation (as long as $a \neq 0$). The term $b^2 - 4ac$, called the discriminant, tells us about the nature of the roots (real and distinct, real and equal, or complex).
- Checking Sum/Product of Coefficients: As demonstrated in this problem, checking the sum of coefficients ($a+b+c$) and the alternating sum ($a-b+c$) can quickly identify if $1$ or $-1$ are roots, respectively.
Understanding these different methods helps in choosing the most efficient way to solve a given quadratic equation.
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