Common Roots: Key Word Parts That Build Vocabulary

When two quadratic equations are considered, a root is termed as common if it satisfies both equations simultaneously. The determination of conditions for existence and evaluation of such roots employs systematic algebraic procedures, foundational formulas, and explicit stepwise reasoning as per NCERT and standard competitive exam practice.

Canonical Quadratic Equation and Its Roots

A general quadratic equation in one variable $x$ is written as $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are real coefficients with $a \neq 0$.

The roots of the given quadratic equation, denoted by $\alpha$ and $\beta$, are values of $x$ where the equation evaluates to zero.

Quadratic Formula: The roots of $ax^2 + bx + c = 0$ are given by: \[ x = \frac{ -b \pm \sqrt{b^2 - 4ac} }{ 2a } \]

Discriminant: The expression $D = b^2 - 4ac$ determines the nature of the roots: If $D > 0$, roots are real and distinct. If $D = 0$, roots are real and equal. If $D < 0$, roots are complex conjugate.

Definition of Common Roots For Two Quadratic Equations

Consider two quadratic equations, \[ p_1 x^2 + q_1 x + r_1 = 0 \tag{1} \] and \[ p_2 x^2 + q_2 x + r_2 = 0 \tag{2} \] with $p_1, p_2 \neq 0$. A common root is a value of $x$ satisfying both (1) and (2) simultaneously.

Explicit Condition for Existence of a Single Common Root

Let $a$ denote the common root. Then $a$ satisfies both equations, so: \begin{align} p_1 a^2 + q_1 a + r_1 &= 0 \quad \tag{3} \\ p_2 a^2 + q_2 a + r_2 &= 0 \quad \tag{4} \end{align}

Subtracting (4) from (3) yields: \[ (p_1 - p_2)a^2 + (q_1 - q_2)a + (r_1 - r_2) = 0 \] However, this does not directly determine $a$. To obtain explicit expressions for $a$, treat (3) and (4) as simultaneous equations in $a^2$ and $a$. Rearranging (3) and (4): \begin{align} p_1 a^2 + q_1 a + r_1 = 0 \\ p_2 a^2 + q_2 a + r_2 = 0 \end{align}

Subtract $p_2$ times (3) from $p_1$ times (4): \[ p_2 (p_1 a^2 + q_1 a + r_1) - p_1 (p_2 a^2 + q_2 a + r_2) = 0 \]

Expanding left-hand side: \begin{align*} &\quad\ p_2 p_1 a^2 + p_2 q_1 a + p_2 r_1 \\ &- p_1 p_2 a^2 - p_1 q_2 a - p_1 r_2 \end{align*}

Grouping terms: \[ (p_2 q_1 - p_1 q_2) a + (p_2 r_1 - p_1 r_2) = 0 \]

Assuming $p_1 q_2 \neq p_2 q_1$, the above yields: \[ a = \frac{p_1 r_2 - p_2 r_1}{p_2 q_1 - p_1 q_2} \] Note that the sign is reversed in numerator for consistency, so rewrite as: \[ a = -\frac{p_1 r_2 - p_2 r_1}{p_1 q_2 - p_2 q_1} \]

To verify $a$ is indeed a root, substitute back into (3): \[ p_1 a^2 + q_1 a + r_1 = 0 \] Substitute the value of $a$: \[ a = -\frac{p_1 r_2 - p_2 r_1}{p_1 q_2 - p_2 q_1} \] Calculate $a^2$ as: \[ a^2 = \left( \frac{p_1 r_2 - p_2 r_1}{p_1 q_2 - p_2 q_1} \right)^2 \]

Upon substitution into $p_1 a^2 + q_1 a + r_1$, the resulting equation leads to a consistency condition: \[ p_1 a^2 + q_1 a + r_1 = 0 \] \[ p_1 \left( \frac{p_1 r_2 - p_2 r_1}{p_1 q_2 - p_2 q_1} \right)^2 + q_1 \left( -\frac{p_1 r_2 - p_2 r_1}{p_1 q_2 - p_2 q_1} \right) + r_1 = 0 \]

Multiply both sides by $(p_1 q_2 - p_2 q_1)^2$ to clear denominators: \begin{align*} p_1 (p_1 r_2 - p_2 r_1)^2 - q_1 (p_1 r_2 - p_2 r_1)(p_1 q_2 - p_2 q_1) + r_1 (p_1 q_2 - p_2 q_1)^2 = 0 \end{align*}

After explicit expansion and careful term-by-term grouping, the ultimate necessary and sufficient condition for the existence of a single common root among two quadratics, where the equations are not scalar multiples of each other, is: \[ (p_1 r_2 - p_2 r_1)^2 = (q_1 r_2 - q_2 r_1)(p_1 q_2 - p_2 q_1) \]

If this condition holds, then exactly one root is common to both equations, and its explicit value is $a = -\frac{p_1 r_2 - p_2 r_1}{p_1 q_2 - p_2 q_1}$, provided $p_1 q_2 \neq p_2 q_1$.

Explicit Condition for Existence of Two Common Roots

When two quadratic equations share both roots, they are proportional; each equation is a scalar multiple of the other. Let their roots be $a, b$. Write: For $p_1 x^2 + q_1 x + r_1 = 0$: \begin{align*} a + b &= -\frac{q_1}{p_1} \\ ab &= \frac{r_1}{p_1} \end{align*} and for $p_2 x^2 + q_2 x + r_2 = 0$: \begin{align*} a + b &= -\frac{q_2}{p_2} \\ ab &= \frac{r_2}{p_2} \end{align*}

Equate the two sets: \begin{align*} -\frac{q_1}{p_1} &= -\frac{q_2}{p_2} \implies \frac{q_1}{p_1} = \frac{q_2}{p_2} \\ \frac{r_1}{p_1} &= \frac{r_2}{p_2} \end{align*} Therefore, the ratios $\frac{p_1}{p_2}$, $\frac{q_1}{q_2}$, and $\frac{r_1}{r_2}$ must be equal.

Necessary and Sufficient Condition: For two quadratics to have both roots common, \[ \frac{p_1}{p_2} = \frac{q_1}{q_2} = \frac{r_1}{r_2} \] Here, the two equations are scalar multiples.

Criterion for Exactly One Root Common

For exactly one root to be common between two quadratics, two conditions must be satisfied:

(i) They are not scalar multiples (i.e., they are not proportional as polynomials).

(ii) The single root criterion derived above, specifically, \[ (p_1 r_2 - p_2 r_1)^2 = (q_1 r_2 - q_2 r_1)(p_1 q_2 - p_2 q_1) \] with $p_1 q_2 - p_2 q_1 \neq 0$.

Necessary and Sufficient Condition for Both Roots Equal (Repeated Common Roots)

If both quadratics have a repeated root in common, i.e., both are equal and common, then each quadratic must have repeated roots, i.e., both discriminants must be zero: \[ (q_1^2 - 4 p_1 r_1) = 0, \quad (q_2^2 - 4 p_2 r_2) = 0 \] and the proportionality condition $\frac{p_1}{p_2} = \frac{q_1}{q_2} = \frac{r_1}{r_2}$ must also hold.

Worked Example: Determining Common Roots Between Two Quadratic Equations

Given: $2x^2 + 3x + 1 = 0$ and $x^2 + 2x + 1 = 0$

Step 1: Compute $p_1 r_2 - p_2 r_1$. \[ p_1 = 2,\ r_1 = 1;\ p_2 = 1,\ r_2 = 1 \] \[ p_1 r_2 - p_2 r_1 = 2 \cdot 1 - 1 \cdot 1 = 2 - 1 = 1 \]

Step 2: Compute $q_1 r_2 - q_2 r_1$. \[ q_1 = 3,\ q_2 = 2 \] \[ q_1 r_2 - q_2 r_1 = 3 \cdot 1 - 2 \cdot 1 = 3 - 2 = 1 \]

Step 3: Compute $p_1 q_2 - p_2 q_1$. \[ p_1 q_2 - p_2 q_1 = 2 \cdot 2 - 1 \cdot 3 = 4 - 3 = 1 \]

Step 4: Check the single common root criterion. \[ (p_1 r_2 - p_2 r_1)^2 = 1^2 = 1 \] \[ (q_1 r_2 - q_2 r_1)(p_1 q_2 - p_2 q_1) = 1 \cdot 1 = 1 \]

Since $1 = 1$, the criterion is satisfied, and exactly one root is common between the two equations.

Step 5: Calculate the common root explicitly. \[ a = -\frac{p_1 r_2 - p_2 r_1}{p_1 q_2 - p_2 q_1} = -\frac{1}{1} = -1 \]

Thus, $x = -1$ is the single common root of both quadratics.

Common roots represent a vital intersection of solutions in the study of quadratic equations, and the derived conditions explained above are foundational for JEE Main algebraic problem solving. For further study on advanced polynomial relationships, readers may review the related topic Properties Of Triangle And Height And Distance.