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# Common Roots of a Quadratic Equation - JEE Important Topic

## A Quadratic Equation in a Nutshell

Last updated date: 20th Mar 2023
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Any algebraic equation could be described as an expression with a combination of variables, numbers, and operations. There are many algebraic equations categorised based on the degree of the expression. One such expression with the second degree, i.e., the highest power of a variablein the equation is 2, is called a quadratic equation. It is named after the Quad that means a shape having four sides, which specifies a square. This is why we read x2 as x square in the quadratic equation.

A general quadratic equation is represented in the form, $ax^2 + bx + c = 0$

where x represents the variable and a, b, and c are called coefficients of the expression and can be named as the quadratic coefficient, linear coefficient and the constant term respectively.

Here, $a\,\ne \,0$ and that is because if there is no quadratic term in the equation which has the power 2, then the equation becomes linear, whereas the equation remains quadratic even if b or c is equal to zero. If all the coefficients are real, then there are either two distinct real solutions, or a single and real double root, or two imaginary complex solutions.

## What is the Root Formula?

The roots of the equation are the various values of x for which the quadratic expression becomes zero. In the graphical representation of a quadratic equation on the XY axis, the roots are found using the values where the parabola cuts the X-axis.

Let us assume the two roots to be $\alpha$ and $\beta$. The nature of the two roots can be found using the discriminant, $D=\sqrt{{{b}^{2}}-4ac}$.

• If D = 0, then the two roots are real and equal.

• If D > 0, then the two roots are real and distinct.

• If D < 0, then the two roots are imaginary.

The formula for finding the roots is given by,

\begin{align} & \alpha ,\beta =\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\ & \\ \end{align}

## What is Meant by “Common Root”?

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

Let’s assume two different quadratic equations, namely

$p_1x^2 + q_1x + r_1 = 0$ and $p_2x^2 + q_2x + r_2 = 0$.

## Condition for One Common Root, ‘$a$’

Assuming a condition that both the equations have a common root known as ‘a‘.

This shows that when ‘a‘is substituted in both the equations, the result would be zero.

So, by substituting ‘a ‘in the place of x, we get

$p_1a^2 + q_1a + r_1 = 0$ and $p_2a^2 + q_2a + r_2 = 0$

By using the crammer’s rule, we can solve the above two equations.

$\dfrac{{{a}^{2}}}{\left| \begin{matrix} {{q}_{1}} & {{r}_{1}} \\ {{q}_{2}} & {{r}_{2}} \\ \end{matrix} \right|}=\dfrac{-a}{\left| \begin{matrix} {{p}_{1}} & {{r}_{1}} \\ {{p}_{2}} & {{r}_{2}} \\ \end{matrix} \right|}=\dfrac{1}{\left| \begin{matrix} {{p}_{1}} & {{q}_{1}} \\ {{p}_{2}} & {{q}_{2}} \\ \end{matrix} \right|}$

We obtain the following steps after solving the determinant parts,

$\dfrac{{{a}^{2}}}{{{q}_{1}}{{r}_{2}}-{{q}_{2}}{{r}_{1}}}=\dfrac{-a}{{{p}_{1}}{{r}_{2}}-{{p}_{2}}{{r}_{1}}}=\dfrac{1}{{{p}_{1}}{{q}_{2}}-{{p}_{2}}{{q}_{1}}}$

Equating the first and second equivalents to the third one, we obtain,

\begin{align} \dfrac{a^{2}}{q_{1} r_{2}-q_{2} r_{1}}=\dfrac{1}{p_{1} q_{2}-p_{2} q_{1}} \text { and } \dfrac{-a}{p_{1} r_{2}-p_{2} r_{1}}=\dfrac{1}{p_{1} q_{2}-p_{2} q_{1}} \\ a^{2}=\dfrac{q_{1} r_{2}-q_{2} r_{1}}{p_{1} q_{2}-p_{2} q_{1}}, a=\dfrac{-\left(p_{1} r_{2}-p_{2} r_{1}\right)}{p_{1} q_{2}-p_{2} q_{1}} \end{align}

Substituting the value of a in a2

\begin{align} &\dfrac{\left(p_{1} r_{2}-p_{2} r_{1}\right)^{2}}{\left(p_{1} q_{2}-p_{2} q_{1}\right)^{2}}=\dfrac{q_{1} r_{2}-q_{2} r_{1}}{p_{1} q_{2}-p_{2} q_{1}} \\ &\text { i.e., }\left(p_{1} r_{2}-p_{2} r_{1}\right)^{2}=\left(q_{1} r_{2}-q_{2} r_{1}\right)\left(p_{1} q_{2}-p_{2} q_{1}\right) \end{align}

Therefore, the obtained equation is stated as the condition for one common root between two quadratic equations.

Note: We can also find the common root by making the coefficient of x square term equal and thereby subtracting both the equations to solve them further.

## Condition for Two Common Roots, ‘a‘ and ‘b’

Let us now find out the conditions for the two common roots between the two quadratic equations and generate the common root formula.

Assuming two common roots to be ‘a‘ and ‘b’, we know that the sum of the roots and the product of the roots can be found through the root formula as shown below.

For the equation, $px^2 + qx + r = 0$

Sum of roots, $a+b=\dfrac{-q+D}{2 p}+\dfrac{-q-D}{2 p}=\dfrac{-q+D-q-D}{2 p}=\dfrac{-q}{p}$

Product of roots, $a \cdot b=\dfrac{-q+D}{2 p} \cdot \dfrac{-q-D}{2 p}=\dfrac{q^{2}-D^{2}}{4 p^{2}}=\dfrac{q^{2}-\left(q^{2}-4 p r\right)}{4 p^{2}}=\dfrac{r}{p}$

Now, let us use these formulae for the quadratic equations which have two

common roots,

$p_1x^2 + q_1x + r_1 = 0$ and $p_2x^2 + q_2x + r_2 = 0$

For the first equation, $a+b=\dfrac{-{{q}_{1}}}{{{p}_{1}}},a.b=\dfrac{{{r}_{1}}}{{{p}_{1}}}$

Similarly, for the second equation,  $a+b=\dfrac{-{{q}_{2}}}{{{p}_{2}}},a.b=\dfrac{{{r}_{2}}}{{{p}_{2}}}$

Since the roots are equal thereby the sum and the product of the roots for both the equations will also be equal.

By equating the sum and product equations, we obtain-

$\dfrac{-{{q}_{1}}}{{{p}_{1}}}=\dfrac{-{{q}_{2}}}{{{p}_{2}}}and\dfrac{{{r}_{1}}}{{{p}_{1}}}=\dfrac{{{r}_{2}}}{{{p}_{2}}}$

Therefore, the condition for two roots in common is given by,

$\dfrac{{{p}_{1}}}{{{p}_{2}}}=\dfrac{{{q}_{1}}}{{{q}_{2}}}and\dfrac{{{r}_{1}}}{{{r}_{2}}}$

## Equal Roots Condition

For a quadratic equation to have equal roots, the discriminant, D should be equal to zero.

I.e.    $D=\sqrt{{{b}^{2}}-4ac}= 0$

If the two quadratic equations have equal and common roots, then

Discriminant of the first quadratic equation = 0 = Discriminant of the second quadratic equation

Let us conclude the article.

## Conclusion

Therefore, the conditions for one and two common roots between two quadratic equations are derived and stated as proved. When two random quadratic equations are given along with a common root condition statement, then start substituting the conditions derived 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.

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## FAQs on Common Roots of a Quadratic Equation - JEE Important Topic

1. What exactly is a Quadratic Equation? How do you determine the nature of the roots through the equation?

A quadratic equation is an algebraic polynomial expression consisting of variables whose highest order is 2. It is represented in the form of ax2 + bx + c = 0

Where a, b and c belong to the set of real numbers and a cannot be equal to zero. We find the nature of the roots using the value of the discriminant. If D > 0, the roots are real and distinct. If D = 0, the roots are real and equal. If D < 0, roots are imaginary or complex.

2. How to find the nature of the roots through the graph of the quadratic equation?

In essence, the graph of a quadratic equation is a parabola. If the parabola cuts the X-axis at two different points, then the two roots are real and distinct, which are the two intercepts on the X-axis. If the parabola makes a tangent with the X-axis, then the roots are equal and real, which is also called a double root, which is the point of contact on the X-axis. If the parabola doesn’t touch the X-axis, then the two roots are assumed as imaginary or complex.