Condition for common roots

A quadratic equation is a second-order polynomial equation in a single variable. It is a second-degree algebraic expression and is of the form ax² + bx + c = 0. In other words, a quadratic equation is an “equation of degree 2.” Roots are the solutions of any type of equation. There are many scenarios where a quadratic equation is used. The value of the discriminant determines the type of solution or roots. Discriminant can be found using the formula: D = b² – 4ac. Further, a quadratic equation has numerous applications in physics, engineering, astronomy, etc.

In this article, we will cover the concept of the common roots of two quadratic equations. This concept falls under the broader category of complex numbers and quadratic equations, a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more.

What is meant by common root?

A common root is a root that satisfies the root properties of two or more equations.

We consider two-equation and try to find the conditions for roots to be common among them

Let the equations be $a_1 x^2+b_1 x+c_1=0$ and $a_2 x^2+b_2 x+c_2=0$
Only one common root:
Let $\alpha$ be the common root, so it will satisfy both the equations
This, $\mathrm{a}_1 \alpha^2+\mathrm{b}_1 \alpha+\mathrm{c}_1=0$ and $\mathrm{a}_2 \alpha^2+\mathrm{b}_2 \alpha+\mathrm{c}_2=0$
On solving these two equations using cross multiplication method, we get

$$
\Rightarrow \alpha^2=\frac{b_1 c_2-b_2 c_1}{a_1 b_2-a_2 b_1}
$$

and,

$$
\Rightarrow \alpha=\frac{\mathrm{c}_1 \mathrm{a}_2-\mathrm{c}_2 \mathrm{a}_1}{\mathrm{a}_1 \mathrm{~b}_2-\mathrm{a}_2 \mathrm{~b}_1}
$$

from above equations, we can write

$$
\begin{aligned}
& \Rightarrow \frac{b_1 c_2-b_2 c_1}{a_1 b_2-a_2 b_1}=\left(\frac{c_1 a_2-c_2 a_1}{a_1 b_2-a_2 b_1}\right)^2 \\
& \Rightarrow\left(b_1 c_2-b_2 \mathbf{c}_1\right)\left(\mathbf{a}_1 \mathbf{b}_2-\mathbf{a}_2 b_1\right)=\left(\mathbf{c}_1 \mathbf{a}_2-\mathbf{c}_2 \mathbf{a}_1\right)^2
\end{aligned}
$$

This is the condition required for a root to be common to both quadratic equations.

And the common root is given by,

$$
\alpha=\frac{b_1 c_2-b_2 c_1}{a_1 b_2-a_2 b_1}=\frac{c_1 a_2-c_2 a_1}{a_1 b_2-a_2 b_1}
$$

The common root can also be found using the method given below :
1. First make the coefficient of $x^2$ the same in two given quadratic equations.
2. Now, subtract the two equations
3. Use the relation between their roots and coefficients to find the roots of other equations.

Both roots are common :
Let $\alpha$ and $\beta$ be the common roots of the equations

$$
a_1 x^2+b_1 x+c_1=0 \text { and } a_2 x^2+b_2 x+c_2=0
$$

Then, both the equation are identical, hence

$$
\frac{\mathbf{a}_1}{\mathbf{a}_2}=\frac{\mathbf{b}_1}{\mathbf{b}_2}=\frac{\mathbf{c}_1}{\mathbf{c}_2}
$$

Equal Roots Condition
For a quadratic equation to have equal roots, the discriminant, $D$ should be equal to zero.
I.e.

$$
\mathrm{D}=b^2-4 a c=0
$$

If the two quadratic equations have equal and common roots, then
Discriminant of the first quadratic equation $=0=$ Discriminant of the second quadratic equation

Solved Examples Based On the Condition of Common Roots:
Example 1: $a x^2+b x+c=0($ Where $a, b, c \epsilon I)$ and $x^2+5 x+2=0$ have a common root then $a: b: c$ equals
Solution:
As we learned in
Condition for both roots common -

$$
\frac{a}{a^{\prime}}=\frac{b}{b^{\prime}}=\frac{c}{c^{\prime}}
$$

- wherein

$$
\begin{aligned}
& a x^2+b x+c=0_{\&} \\
& a^{\prime} x^2+b^{\prime} x+c^{\prime}=0
\end{aligned}
$$

are the 2 equations
$x^2+5 x+2=0$ has irrational roots and both equations one given to have a common root. Since $a, b, c \in Q$ so $a x^2+b x+c=0$ to satisfy a condition of a common root, it will have both roots common.

$$
\text { So, } \frac{a}{1}=\frac{b}{5}=\frac{c}{2} \Rightarrow a: b: c=1: 5: 2
$$

Hence, the answer is 1:5:2.

Example 2: $a x^2+b x+c=0$ and $x^2+2 x+5=0$ have both roots are common. Then minimum possible value of $(a+b+c)($ Where $a, b, c \in N)$ is

Solution:
As we learned in
Condition for both roots common -

$$
\frac{a}{a^{\prime}}=\frac{b}{b^{\prime}}=\frac{c}{c^{\prime}}
$$

- wherein

$$
\begin{aligned}
& a x^2+b x+c=0_{\&} \\
& a^{\prime} x^2+b^{\prime} x+c^{\prime}=0
\end{aligned}
$$

are the 2 equations
$\because$ Both roots are common so, $\frac{a}{1}=\frac{b}{2}=\frac{c}{5}$
for minimum $a+b+c$ :

$$
a=1, b=2 \text { and } c=5
$$

Hence, the answer is 8 .

Example 3: Which of the following is the condition of exactly one root common between $a x^2+b x+c=0$ and $c x^2+b x+a=0$
Solution:
As we learned in
Condition for one common root -

$$
\left(a^{\prime} c-a c^{\prime}\right)^2=\left(b c^{\prime}-b^{\prime} c\right)\left(a b^{\prime}-a^{\prime} b\right)
$$

- wherein

$$
\begin{aligned}
& a x^2+b x+c=0_{\&} \\
& a^{\prime} x^2+b^{\prime} x+c^{\prime}=0
\end{aligned}
$$

are the 2 equations

$$
\begin{aligned}
& a^2+b x+c=0 \\
& c x^2+b x+a=0 \\
& \Rightarrow\left(a^2-c^2\right)^2=(a b-b c)(a b-b c) \\
& \Rightarrow(a-c)^2(a+c)^2=b^2(a-c)^2 \\
& \Rightarrow(a-c)^2\left[(a+c)^2-b^2\right]=0 \\
& \Rightarrow a=c \text { or } a+b+c=0 \quad \text { or } a-b+c=0
\end{aligned}
$$

When $a=c$ then both quadratics are the same and have both roots are common.
$a+b+c=0$ will give $x=1$ as common root.

$a-b+c=0$ will give $x=-1$ as common root.
In the above options $(A) \&(B)$ are only conditions of common roots in which $(A)$ will be rejected as explained above.
Hence, the answer is the option 2.
Example 4: If $a x^2+b x+c=0$ and $b x^2+c x+a=0$ have a common root and $a, b, c$ are non-zero real numbers, then $\frac{a^3+b^3+c^3}{a b c}$ equals
Solution:
As we learned in
Condition for one common root -

$$
\left(a^{\prime} c-a c^{\prime}\right)^2=\left(b c^{\prime}-b^{\prime} c\right)\left(a b^{\prime}-a^{\prime} b\right)
$$

- wherein

$$
\begin{aligned}
& a x^2+b x+c=0 \& \\
& a^{\prime} x^2+b^{\prime} x+c^{\prime}=0
\end{aligned}
$$

are the 2 equations
$a x^2+b x+c=0$ and $b x^2+c x+a=0$ have common roots, so
$\left(a^2-b c\right)^2=\left(a c-b^2\right)\left(a b-c^2\right)$
$\Rightarrow a^4+b^2 c^2-2 a^2 b c=a^2 b c-a c^3-a b^3+b^2 c^2$
$\Rightarrow a^4+a b^3+a c^3=3 a^2 b c$
$\Rightarrow a^3+b^3+c^3=3 a b c$

Hence, the answer is 3.

Example 5: The value of ' $a$ ' for which $x^2-11 x+a=0$ and $x^2-14 x+2 a=0$ have common roots are
Solution:
As we learned in
Condition for one common root -

$$
\left(a^{\prime} c-a c^{\prime}\right)^2=\left(b c^{\prime}-b^{\prime} c\right)\left(a b^{\prime}-a^{\prime} b\right)
$$

- wherein

$$
\begin{aligned}
& a x^2+b x+c=0 \& \\
& a^{\prime} x^2+b^{\prime} x+c^{\prime}=0
\end{aligned}
$$

are the 2 equations
$x^2-11 x+a=0$ and $x^2-14 x+2 a=0$ have a common root.

$$
\begin{aligned}
& \therefore(1 \times a-1 \times 2 a)^2=[(-11)(2 a)-(-14)(a)] \times[1 \times(-14)-1(-11)] \\
& \Rightarrow a^2=(-8 a)(-3) \\
& \Rightarrow a^2-24 a=0 \Rightarrow a=0 \text { or } a=24
\end{aligned}
$$
Hence, the answer is 0 or 24 .

Summary

Summary here to prove or find the roots. So, hereby inferring that the common root formula and common root conditions are useful in the calculations of common roots of quadratic equations.