If two equations ${x^2} - cx + d = 0$ and ${x^2} - ax + b = 0$ have one common root and the second has equal roots, then $2(b + d) = $
A. $0$
B. $a + c$
C. $ac$
D. $ - ac$
Answer
301.5k+ views
Hint: We have to establish a relation between coefficients of the given quadratic equations. To proceed further we have to find the sum and product of the roots of the given quadratic equations. For the first equation we will simply find the sum and product of the roots. But according to the question, the second equation has equal roots. Taking this under consideration, we will find the sum and product of the equal roots too. And at last, substitute all these values accordingly to get the result.
Formula Used: For a quadratic equation $a{x^2} + bx + c = 0$ having roots $\alpha ,\beta $ . The sum of roots is: $(\alpha + \beta ) = - \dfrac{b}{a}$ and the product of roots is: $(\alpha \beta ) = \dfrac{c}{a}$ .
Complete step-by-step solution:
We have two quadratic equations
${x^2} - cx + d = 0$ ----(1.1)
${x^2} - ax + b = 0$ ----(1.2)
If we compare equations (1.1) and (1.2) with ${a_1}{x^2} + {b_1}x + {c_1} = 0$ and ${a_2}{x^2} + {b_2}x + {c_2} = 0$, we will get ${a_1} = 1$ , ${b_1} = - c$ , ${c_1} = d$ ; ${a_2} = 1$ , ${b_2} = - a$ , ${c_2} = b$ .
Suppose, the given equation (1.1) has roots $\alpha ,\beta $ .
Then, the sum of roots will be:
$(\alpha + \beta ) = - \dfrac{{{b_1}}}{{{a_1}}}$
$ \Rightarrow (\alpha + \beta ) = c$
And the product of roots:
$(\alpha \beta ) = \dfrac{{{c_1}}}{{{a_1}}}$
$ \Rightarrow (\alpha \beta ) = d$
Similarly, equation (1.2) has the equal roots, say, $\alpha ,\alpha $ .
$\therefore $ The sum of roots
$ \Rightarrow (\alpha + \alpha ) = 2\alpha = a$
And the product of the roots will be
$ \Rightarrow (\alpha \cdot \alpha ) = {\alpha ^2} = b$
Now, to find the value of $2(b + d)$ , substitute the values of $b$ and $d$ into it, we get
$ \Rightarrow 2(b + d) = 2({\alpha ^2} + \alpha \beta )$
$ \Rightarrow 2(b + d) = 2\alpha (\alpha + \beta )$
We have $2\alpha = a$ and $(\alpha + \beta ) = c$
So,
$ \Rightarrow 2(b + d) = ac$
Hence, the correct option is C.
Note: As it is given in the question that the equations have one common root. So, we can also use the method of cramer’s rule to get the value of the common root. And then, we can proceed with the process of finding the sum of the roots and product of the roots. And the second equation has equal roots, so the process for this will be the same.
Formula Used: For a quadratic equation $a{x^2} + bx + c = 0$ having roots $\alpha ,\beta $ . The sum of roots is: $(\alpha + \beta ) = - \dfrac{b}{a}$ and the product of roots is: $(\alpha \beta ) = \dfrac{c}{a}$ .
Complete step-by-step solution:
We have two quadratic equations
${x^2} - cx + d = 0$ ----(1.1)
${x^2} - ax + b = 0$ ----(1.2)
If we compare equations (1.1) and (1.2) with ${a_1}{x^2} + {b_1}x + {c_1} = 0$ and ${a_2}{x^2} + {b_2}x + {c_2} = 0$, we will get ${a_1} = 1$ , ${b_1} = - c$ , ${c_1} = d$ ; ${a_2} = 1$ , ${b_2} = - a$ , ${c_2} = b$ .
Suppose, the given equation (1.1) has roots $\alpha ,\beta $ .
Then, the sum of roots will be:
$(\alpha + \beta ) = - \dfrac{{{b_1}}}{{{a_1}}}$
$ \Rightarrow (\alpha + \beta ) = c$
And the product of roots:
$(\alpha \beta ) = \dfrac{{{c_1}}}{{{a_1}}}$
$ \Rightarrow (\alpha \beta ) = d$
Similarly, equation (1.2) has the equal roots, say, $\alpha ,\alpha $ .
$\therefore $ The sum of roots
$ \Rightarrow (\alpha + \alpha ) = 2\alpha = a$
And the product of the roots will be
$ \Rightarrow (\alpha \cdot \alpha ) = {\alpha ^2} = b$
Now, to find the value of $2(b + d)$ , substitute the values of $b$ and $d$ into it, we get
$ \Rightarrow 2(b + d) = 2({\alpha ^2} + \alpha \beta )$
$ \Rightarrow 2(b + d) = 2\alpha (\alpha + \beta )$
We have $2\alpha = a$ and $(\alpha + \beta ) = c$
So,
$ \Rightarrow 2(b + d) = ac$
Hence, the correct option is C.
Note: As it is given in the question that the equations have one common root. So, we can also use the method of cramer’s rule to get the value of the common root. And then, we can proceed with the process of finding the sum of the roots and product of the roots. And the second equation has equal roots, so the process for this will be the same.
Recently Updated Pages
If a parabola whose length of latus rectum is 4a touches class 11 maths JEE_Main

Find the cubic polynomial whose zeroes are 3 5 and class 11 maths JEE_Main

During the sale colour pencils were being sold in -class-11-maths-JEE_Main

A man on the top of a vertical observation tower o-class-11-maths-JEE_Main

In a class of 60 students 25 students play cricket class 11 maths JEE_Main

A regular polygon has 20 sides How many triangles can class 11 maths JEE_Main

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Electron Gain Enthalpy and Electron Affinity Explained

Derivation of Equation of Trajectory Explained for Students

Understanding Atomic Structure for Beginners

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
NCERT Solutions For Class 11 Maths Chapter 6 Permutations And Combinations - 2026-27 Free PDF Download (Login Required)

NCERT Solutions For Class 11 Maths Chapter 9 Straight Lines - 2026-27 Free PDF Download (Sign-in Required)

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

NCERT Solutions For Class 11 Maths Chapter 8 Sequences And Series - 2026-27 Free PDF Download (Login Required)

NCERT Solutions For Class 11 Maths Chapter 4 Complex Numbers And Quadratic Equations - 2026-27 Free PDF Download (Login Required)

Hybridisation in Chemistry – Concept, Types & Applications

