Common Roots: A polynomial is an expression made up of variables and constants in which the exponents of the variables are only positive integers. The polynomial terms are separated mainly by either addition or subtraction operators. Polynomials find their applications in writing polynomial equations and defining polynomial functions.

An example of a polynomial expression is \({x^2} + 3x – 4.\) In this polynomial, the degree is \(2\) since the term with the highest power of \(x\) is \({x^2}\) This polynomial has three terms. Polynomials can be categorised with respect to their total number of terms and their degree. 

Types of Polynomials Based on Degree

The power of the leading term or the highest power of the variable is called the degree of the polynomial. This is obtained by arranging the polynomial terms in the descending order of their powers. Based on the degree of the polynomial, they can be classified into four major types. They are,

• Zero or constant polynomial
• Linear polynomial
• Quadratic polynomial
• Cubic polynomial

The roots of equations are also known as the ‘solutions’ or ‘zeros’ of the equation. There are different ways to solve the equations, such as factoring, completing the square, or graphing. So, can equations have common roots or distinct roots only?

Let us understand the concept of common roots in this article and some solved examples.

What are Roots?

The roots of quadratic equations are nothing but solutions to that equation. They are the values of the variables that satisfy the given equation. And the number of roots of an equation is equal to the degree of that equation.

For example, since the degree of the quadratic equation is \(2,\) it will have a maximum of two roots.

Similarly, while comparing the roots of quadratic equations, they may have one or more roots in common.

Conditions for Common Root

The condition for common roots of quadratic equations is listed below.

Condition for one Common Root

Let two quadratic equations be

\({a_1}{x^2} + {b_1}x + {c_1} = 0\)

\({a_2}{x^2} + {b_2}x + {c_2} = 0\)

Let us find the condition that the above quadratic equations may have a common root.

Let \(\alpha \) be the common root of the given quadratic equations.

Then,

\({a_1}{\alpha ^2} + {b_1}\alpha + {c_1} = 0\)

\({a_2}{\alpha ^2} + {b_2}\alpha + {c_2} = 0\)

By Crammer’s rule,

\(\frac{{{\alpha ^2}}}{{\left| {\begin{array}{*{20}{l}} { – {c_1}}&{{b_1}}\\ { – {c_2}}&{{b_2}} \end{array}} \right|}} = \frac{\alpha }{{\left| {\begin{array}{*{20}{l}} {{a_1}}&{ – {c_1}}\\ {{a_2}}&{ – {c_2}} \end{array}} \right|}} = \frac{1}{{\left| {\begin{array}{*{20}{l}} {{a_1}}&{{b_1}}\\ {{a_2}}&{{b_2}} \end{array}} \right|}}\)

\(\Rightarrow \frac{{{\alpha ^2}}}{{{b_1}{c_2} – {b_2}{c_1}}} = \frac{\alpha }{{{a_2}{c_1} – {a_1}{c_2}}} = \frac{1}{{{a_1}{b_2} – {a_2}{b_1}}}\)

\(\therefore \alpha = \frac{{{a_2}{c_1} – {a_1}{c_2}}}{{{a_1}{b_2} – {a_2}{b_1}}}\)

\(\alpha = \frac{{{b_1}{c_2} – {b_2}{c_1}}}{{{a_2}{c_1} – {a_1}{c_2}}},\alpha \ne 0\)

Hence, the condition for one root is common is \({\left( {{c_1}{a_2} – {c_2}{a_1}} \right)^2} = \left( {{b_1}{c_2} – {b_2}{c_1}} \right)\left( {{a_1}{b_2} – {a_2}{b_1}} \right).\)

We can also write the common root as

\(\alpha = \frac{{{a_2}{c_1} – {a_1}{c_2}}}{{{a_1}{b_2} – {a_2}{b_1}}},\alpha \ne 0\)

or

\(\alpha = \frac{{{b_1}{c_2} – {b_2}{c_1}}}{{{a_2}{c_1} – {a_1}{c_2}}},\alpha \ne 0\)

Note:

We can find the common root by

• making the same coefficient of \({x^2}\) of the given quadratic equations and then subtract them.
• using the relation between roots and coefficients of the quadratic equation.

Condition for Both Roots Common

Consider the two quadratic equations:

\({a_1}{x^2} + {b_1}x + {c_1} = 0\)

\({a_2}{x^2} + {b_2}x + {c_2} = 0\)

Let \(\alpha ,\beta \) be the common roots of the given quadratic equations.

Therefore,

\(\alpha + \beta = – \frac{{{b_1}}}{{{a_1}}}\)

\(\alpha \beta = \frac{{{c_1}}}{{{a_1}}}\)

\(\alpha + \beta = – \frac{{{b_2}}}{{{a_2}}}\)

\(\alpha \beta = \frac{{{c_2}}}{{{a_2}}}\)

Thus, \( – \frac{{{b_1}}}{{{a_1}}} = – \frac{{{b_2}}}{{{a_2}}}\) and \(\frac{{{c_1}}}{{{a_1}}} = \frac{{{c_2}}}{{{a_2}}}\)

\( \Rightarrow \frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}}\) and \(\frac{{{a_1}}}{{{a_2}}} = \frac{{{c_1}}}{{{c_2}}}\)

\( \Rightarrow \frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}\) is the required condition for the both roots are common for two quadratic equations

Note:

Two different quadratic equations with rational coefficients cannot have a single common root that is complex or irrational, as imaginary and surd roots always occur in conjugate pairs.

Cubic Roots

The general form of a cubic form is \(f(x) = a{x^3} + b{x^2} + c{x^1} + d\) and the cubic equation has the form \(a{x^3} + b{x^2} + cx + d = 0,\) where \(a,b\)  and \(c\) are coefficients, and \(d\) is a constant.

How to Solve Cubic Equations?

The most general way of solving a cubic equation is to reduce it into a quadratic equation and then solve it using factorisation or the quadratic formula.

Like a quadratic equation has two real roots, a cubic equation may have possibly three real roots. But, unlike a quadratic equation, which may have no real solution, a cubic equation has at least one real root.

Relation Between Coefficients and Roots

For a cubic equation \(a{x^3} + b{x^2} + cx + d = 0,\) let \(p,q\) and \(r\) be its roots, then the relation between coefficients and roots are given

Expression	Equals to
\(p + q + r\)	\( – \frac{b}{a}\)
\(pq + qr + rp\)	\(\frac{c}{a}\)
\(pqr\)	\( – \frac{d}{a}\)

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Solved Examples on Common Roots

Q.1. If the equations \(3{x^2} – 2x + p = 0\) and \(6{x^2} – 17x + 12 = 0\) have a common root, then find the value of \(p.\)
Ans:
Given: \(3{x^2} – 2x + p = 0\) and \(6{x^2} – 17x + 12 = 0\)
Let \(\alpha \) be the common root of \(3{x^2} – 2x + p = 0\) and \(6{x^2} – 17x + 12 = 0\)
Then \(\alpha \) will satisfy the both equations
Then \(3{\alpha^2} – 2\alpha + p = 0……\left( {\rm{i}} \right)\) and \(6{\alpha^2} – 17\alpha + 12 = 0…..\left( {{\rm{ii}}} \right)\)
Now, multiply equation \(\left( {\rm{i}} \right)\) with \(2\) and then subtract from the equation \(\left( {\rm{ii}} \right),\) then \( – 13\alpha + 12 – 2p = 0\)
\( \Rightarrow 13\alpha – 12 + 2p = 0\)
\( \Rightarrow 13\alpha = 12 – 2p\)
\( \Rightarrow \alpha = \frac{{12 – 2p}}{{13}}\)
Since \(\alpha \) is a common root of the equations, \(3{x^2} – 2x + p = 0\) and \(6{x^2} – 17x + 12 = 0\).
Then, \(3{\alpha ^2} – 2\alpha + p = 0\)
\( \Rightarrow 3{\left( {\frac{{12 – 2p}}{{13}}} \right)^2} – 2\left( {\frac{{12 – 2p}}{{13}}} \right) + p = 0\)
\( \Rightarrow \frac{{3{{(12 – 2p)}^2}}}{{169}} – 2\left( {\frac{{12 – 2p}}{{13}}} \right) + p = 0\)
\( \Rightarrow 3{(12 – 2p)^2} – 26(12 – 2p) + 169p = 0\)
\( \Rightarrow 3\left( {144 – 48p + 4{p^2}} \right) – 312 + 52p + 169p = 0\)
\( \Rightarrow 432 – 144p + 12{p^2} – 312 + 52p + 169p = 0\)
\( \Rightarrow 12{p^2} + 77p + 120 = 0\)
\( \Rightarrow 12{p^2} + 45p + 32p + 120 = 0\)
\( \Rightarrow 3p(4p + 15) + 8(4p + 15) = 0\)
\( \Rightarrow (3p + 8)(4p + 15) = 0\)
\(\therefore p = – \frac{8}{3}, – \frac{{15}}{4}.\)

Q.2. If \(a,b,c \in R\) and the equations \(a{x^2} + bx + c = 0\) and \({x^2} + 2x + 9 = 0,\) have a common root, then show that \(a:b:c = 1:2:9.\)
Ans: Given: \(a{x^2} + bx + c = 0\) and \({x^2} + 2x + 9 = 0\)
Now, discriminant for \({x^2} + 2x + 9 = 0,\) is \(D = {(2)^2} – 4(9)(1)\)
\( \Rightarrow D = 4 – 36\)
\( \Rightarrow D = – 32 < 0\)
Thus, the discriminant is less than zero. So, the roots of the quadratic equation \({x^2} + 2x + 9 = 0\) are imaginary.
As the imaginary roots of a quadratic equation are conjugate, both the roots of the equations will be common.
Therefore, \(\frac{a}{1} = \frac{b}{2} = \frac{c}{9}\)
Or \(a:b:c = 1:2:9\)
Hence, proved.

Q.3. If the roots of \(k\left( {6{x^2} + 3} \right) + rx + 2{x^2} – 1 = 0\) and \(6k\left( {2{x^2} + 1} \right) + px + 4{x^2} – 2 = 0\) are common, then find the value of \(2r – p.\)
Ans: Given: \(k\left( {6{x^2} + 3} \right) + rx + 2{x^2} – 1 = 0\) and \(6k\left( {2{x^2} + 1} \right) + px + 4{x^2} – 2 = 0\)
So, given equations can be written as,
\((6k + 2){x^2} + rx + 3k – 1 = 0\)
\(2(6k + 2){x^2} + px + 2(3k – 1) = 0\)
Since both roots are common, \(\frac{{12k + 4}}{{6k + 2}} = \frac{p}{r} = \frac{{6k – 2}}{{3k – 1}}\)
\( \Rightarrow \frac{p}{r} = 2\)
\( \Rightarrow p = 2r\)
\(\therefore 2r – p = 0\)

Q.4. If the equations \({x^2} – px – 21 = 0,{x^2} – 3px + 35 = 0(p > 0)\) have a common root, then find the value of \(p.\)
Ans: Given: \({x^2} – px – 21 = 0,{x^2} – 3px + 35 = 0\)
Let \(\alpha \) be the common root of \({x^2} – px – 21 = 0,{x^2} – 3px + 35 = 0\)
Then, \(\alpha \) will satisfy both equations.
Then, \({\alpha ^2} – p\alpha – 21 = 0 \ldots {\rm{(i)}}\) and \({\alpha ^2} – 3p\alpha + 35 = 0 \ldots {\rm{(ii)}}\)
Now, subtract the equation \({\rm{(ii)}}\) from the equation \({\rm{(i)}}\),
Then, \(2p\alpha – 56 = 0\)
\( \Rightarrow p\alpha = 28\)
\(\therefore \alpha = \frac{{28}}{p}\)
is a common root of the equations, \({x^2} – px – 21 = 0\) and \({x^2} – 3px + 35 = 0\).
Then, \({\alpha ^2} – p\alpha – 21 = 0\)
\( \Rightarrow {\left( {\frac{{28}}{p}} \right)^2} – p\left( {\frac{{28}}{p}} \right) – 21 = 0\)
\( \Rightarrow \left( {\frac{{784}}{{{p^2}}}} \right) – 28 – 21 = 0\)
\( \Rightarrow 49{p^2} = 784\)
\( \Rightarrow {p^2} = 16\)
\( \Rightarrow p = \pm 4\)
\(\therefore p = 4\) [Since \(p > 0\)]

Q.5. If the equations \({x^2} + px + 10 = 0\) and \({x^2} + qx – 10 = 0\) have a common root, then find the value of \({p^2} – {q^2}.\)
Ans: Given: \({x^2} + px + 10 = 0\) and \({x^2} + qx – 10 = 0\)
Let \(\alpha \) be the common root of \({x^2} + px + 10 = 0\) and \({x^2} + qx – 10 = 0\)
Then, \(\alpha \) will satisfy both equations
Therefore, \({\alpha ^2} + p\alpha + 10 = 0 \ldots {\rm{(i)}}\) and \({\alpha ^2} + q\alpha – 10 = 0 \ldots {\rm{(ii)}}\)
Now, subtract the equation \({\rm{(ii)}}\) from the equation \({\rm{(i),}}\) then \(p\alpha – q\alpha + 20 = 0\)
\( \Rightarrow (p – q)\alpha = – 20\)
\( \Rightarrow \alpha = \frac{{20}}{{q – p}}\)
Since is a common root of the equations, \( {x^2} + px + 10 = 0\) and \({x^2} + qx – 10 = 0.\)
Then, \({\alpha ^2} + p\alpha + 10 = 0\)
\( \Rightarrow {\left( {\frac{{20}}{{q – p}}} \right)^2} + p\left( {\frac{{20}}{{q – p}}} \right) + 10 = 0\)
\( \Rightarrow 400 + 20p(q – p) + 10{(q – p)^2} = 0\)
\( \Rightarrow 400 + 20pq – 20{p^2} + 10{q^2} – 20pq + 10{p^2} = 0\)
\( \Rightarrow 400 – 10{p^2} + 10{q^2} = 0\)
\(\therefore {p^2} – {q^2} = 40\)

Q.6. If the equations \({x^2} + px + q = 0\) and \({x^2} + px + q = 0\) have a common root and \(p \ne q,\) then prove that \(p + q = – 1.\)
Ans: Given: \({x^2} + px + q = 0\) and \({x^2} + px + q = 0\)
To find the common root, given quadratic equations must have the same coefficient of \({x^2}\)
By looking at the given quadratic equations, we can see both equations have the same coefficient.
Let \(\alpha \) be the common root of \({x^2} + px + q = 0\) and \({x^2} + px + q = 0.\)
Then, \({\alpha ^2} + p\alpha + q = 0 \ldots {\rm{(i)}}\) and \({\alpha ^2} + p\alpha + q = 0 \ldots {\rm{(ii)}}\)
Now, subtract equation \({\rm{(ii)}}\) from \({\rm{(i)}}\), then \(\alpha (p – q) + (q – p) = 0\)
\(\alpha (p – q) – (p – q) = 0\)
\((\alpha – 1)(p – q) = 0\)
\(\alpha = 1\) [Since \(p \ne q\)]
Since \(\alpha \) is a common root of the equations, \({x ^2} + px + q = 0\) and \({x^2} + px + q = 0.\)
Then, \({\alpha ^2} + p\alpha + q = 0\)
\(1 + p(1) + q = 0\)
\( \Rightarrow 1 + p + q = 0\)
\( \Rightarrow p + q = – 1\)
Hence, proved.

Q.7. If the quadratic equations \(a{x^2} + 2cx + b = 0\) and \(a{x^2} + 2bx + c = 0(b \ne c)\) have a common root, then prove that \(a + 4b + 4c = 0.\)
Ans: Given: \(a{x^2} + 2cx + b = 0\) and \(a{x^2} + 2bx + c = 0\)
To find the common root, given quadratic equations must have the same coefficient for \({x^2}.\)
Looking at the given two quadratic equations, we can see both equations have the same coefficient.
Let \(\alpha \) be the common root of \(a{x^2} + 2cx + b = 0\) and \(a{x^2} + 2bx + c = 0.\)
Then \(a{\alpha ^2} + 2c\alpha + b = 0 \ldots {\rm{(i)}}\) and \(a{\alpha ^2} + 2b\alpha + c = 0 \ldots {\rm{(ii)}}\)
Now, subtract equation \({\rm{(ii)}}\) from the equation \({\rm{(i),}}\) then \(2\alpha (c – b) + (b – c) = 0\)
\( \Rightarrow 2\alpha (c – b) – (c – b) = 0\)
\( \Rightarrow (2\alpha – 1)(c – b) = 0\)
\( \Rightarrow 2\alpha = 1\) [Since \(b \ne c\)]
\( \Rightarrow \alpha = \frac{1}{2}\)
Since, is a common root of the equations, \(a{x^2} + 2cx + b = 0\) and \(a{x^2} + 2bx + c = 0.\)
Then, \(a{\alpha ^2} + 2c\alpha + b = 0\)
\( \Rightarrow a{\left( {\frac{1}{2}} \right)^2} + 2c\left( {\frac{1}{2}} \right) + b = 0\) [since, \(\alpha = \frac{1}{2}\)]
\( \Rightarrow \frac{a}{4} + c + b = 0\)
\( \Rightarrow a + 4c + 4b = 0\)
Hence, proved.

Q.8. If the equations \(2{x^2} + x + k = 0\) and \({x^2} + \frac{x}{2} – 1 = 0\) have two common roots, then find the value \(k.\)
Ans: Given: \(2{x^2} + x + k = 0\) and \({x^2} + \frac{x}{2} – 1 = 0\)
It is also given that equations \(2{x^2} + x + k = 0\) and \({x^2} + \frac{x}{2} – 1 = 0\) have two roots in common.
As we know, the condition for two roots in common for the quadratic equations \({a_1}{x^2} + {b_1}x + {c_1} = 0\) and \({a_2}{x^2} + {b_2}x + {c_2} = 0\) is \(\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}\)
Therefore, \(\frac{2}{1} = \frac{1}{{\frac{1}{2}}} = \frac{k}{{ – 1}}\)
\( \Rightarrow 2 = 2 = \frac{k}{{ – 1}}\)
\(\therefore k = – 2\)

Q.9. Solve the following cubic equation \({x^3} – 2{x^2} – x + 2 = 0\)
Ans: Given that, we have \({x^3} – 2{x^2} – x + 2 = 0\)
Factorize the equation, then \({x^3} – 2{x^2} – x + 2 = 0\)
\( \Rightarrow {x^2}(x – 2) – 1(x – 2) = 0\)
\( \Rightarrow \left( {{x^2} – 1} \right)(x – 2) = 0\)
\( \Rightarrow (x + 1)(x – 1)(x – 2) = 0\)
\( \Rightarrow x = 1, – 1\) and \(2.\)

Summary

The roots of quadratic equations are nothing but solutions of that equation. They are the values of the variables that satisfy the given equation. Common root is a root which is common to both the given equations. We derived the conditions for one root is common and both roots common. We also covered a few of examples based on common roots i.e., when one root is common to the equations \({a_1}{x^2} + {b_1}x + {c_1} = 0\) and \({a_2}{x^2} + {b_2}x + {c_2} = 0\) and two roots common to the equations \({a_1}{x^2} + {b_1}x + {c_1} = 0\) and \({a_2}{x^2} + {b_2}x + {c_2} = 0.\) Later we explained how to solve cubic roots by using different methods.

FAQs on Common Roots

Q.1. What do you mean by common roots?
Ans: In Mathematics, the common root is a root that is common to the given equations. In other words, the root satisfies the given equations simultaneously.

Q.2. How do you find common roots?
Ans: Let be the common root of quadratic equations \({a_1}{x^2} + {b_1}x + {c_1} = 0\) and \({a_2}{x^2} + {b_2}x + {c_2} = 0.\) Then the condition to find the common root is as \(\alpha = \frac{{{a_2}{c_1} – {a_1}{c_2}}}{{{a_1}{b_2} – {a_2}{b_1}}}\) or \(\alpha = \frac{{{b_1}{c_2} – {b_2}{c_1}}}{{{a_2}{c_1} – {a_1}{c_2}}},\alpha \ne 0\)
Similarly, the condition for both roots common is \(\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}\)

Q.3. What are the types of real roots?
Ans: There are two types of roots:
1. Real and equal roots
2. Real and distinct roots

Q.4. What if two quadratic equations have a common root?
Ans: If two quadratic equations with real coefficients have a non-real complex common root, then both will be common, i.e., both equations will be the same. So the coefficients of the corresponding powers of \(x\) will have proportional values.

Q.5. What is a real root?
Ans: A real root is a root of the quadratic equation \(a{x^2} + bx + c = 0,\) which belong to the set of real numbers. In other words, a real root is a solution to an equation that is also a real number.

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