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
300.9k+ 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
Electron Gain Enthalpy and Electron Affinity Explained

Understanding Atomic Structure for Beginners

Effective Nuclear Charge for JEE

Understanding Inertial and Non-Inertial Frames of Reference

Understanding How a Current Loop Acts as a Magnetic Dipole

Chemistry Formulas List for JEE, NEET & Board Exams - PDF Download (Login Required)

Other Pages
JEE Advanced Weightage Chapter Wise 2026 for Physics, Chemistry, and Mathematics

NCERT Solutions For Class 11 Maths In Hindi Chapter 1 Sets - 2026-27 Free PDF Download (Login Required)

NCERT Solutions For Class 11 Maths Chapter 12 Limits And Derivatives - 2026-27 Free PDF Download (Login Required)

CBSE Notes Class 11 Maths Chapter 4 - Complex Numbers and Quadratic Equations - 2026-27 Free PDF Download (Sign-in Required)

JEE Advanced 2027 Notes

CBSE Notes Class 11 Maths Chapter 6 - Permutations and Combinations - 2026-27 Free PDF Download (Sign-in Required)

