
The equation formed by decreasing each root of \[a{x^2} + bx + c = 0\] by \[1\] is \[2\begin{array}{*{20}{c}}{{x^2} + 8x + 2}& = &0\end{array}\], then
A. \[\begin{array}{*{20}{c}}a& = &{ - b}\end{array}\]
B. \[\begin{array}{*{20}{c}}b& = &{ - c}\end{array}\]
C. \[\begin{array}{*{20}{c}}c& = &{ - a}\end{array}\]
D. \[\begin{array}{*{20}{c}}b& = &{a + c}\end{array}\]
Answer
163.5k+ views
Hint: In this question, first of all, we will determine the sum and the product of the roots of the equation\[\begin{array}{*{20}{c}}{a{x^2} + bx + c}& = &0\end{array}\]. After that, we will do the same for the equation \[2\begin{array}{*{20}{c}}{{x^2} + 8x + 2}& = &0\end{array}\] and from this, we will get that the sum and the product of the roots of the equation are equal. And then we will satisfy the condition from the previous equation and hence, we will get a suitable answer.
Formula Used:
Root’s sum can be determined by
\[\alpha + \beta = \frac{{ - b}}{a}\]
Root’s product can be determined by
\[\alpha \beta = \frac{c}{a}\]
Complete step by step solution:
We have been provided in the question that,
The equation formed by decreasing each root of \[a{x^2} + bx + c = 0\] by \[1\] is
\[2{x^2} + 8x + 2 = 0\]
Let us assume that the roots of the above equation are \[\alpha \]and \[\beta \]respectively.
We know the sum and the product of the roots of the above equation.
Therefore, we can write
\[\begin{array}{*{20}{c}}{ \Rightarrow \alpha + \beta }& = &{ - \frac{b}{a}}\end{array}\] -------- (1)
And
\[\begin{array}{*{20}{c}}{ \Rightarrow \alpha \beta }& = &{\frac{c}{a}}\end{array}\] ------------ (2)
Now we have given that when the roots of the above equation are decreased by \[1\], then the equation \[2\begin{array}{*{20}{c}}{{x^2} + 8x + 2}& = &0\end{array}\]is formed.
Therefore, the new roots will be \[\alpha - 1\]and \[\beta - 1\]respectively.
Now
The sum and the product of the roots of the formed equation will be
\[ \Rightarrow \begin{array}{*{20}{c}}{\left( {\alpha - 1} \right) + \left( {\beta - 1} \right)}& = &{ - \frac{8}{2}}\end{array}\]
\[\begin{array}{*{20}{c}}{ \Rightarrow \alpha + \beta }& = &{ - 2}\end{array}\]
And
\[ \Rightarrow \begin{array}{*{20}{c}}{\left( {\alpha - 1} \right)\left( {\beta - 1} \right)}& = &{\frac{2}{2}}\end{array}\]
\[ \Rightarrow \begin{array}{*{20}{c}}{\alpha \beta - \left( {\alpha + \beta } \right) + 1}& = &1\end{array}\]
\[ \Rightarrow \begin{array}{*{20}{c}}{\alpha \beta }& = &{\alpha + \beta }\end{array}\] -------- (3)
Now put the value of the value of the equation (1) and (2) in the equation (3). Therefore, we will get
\[ \Rightarrow \begin{array}{*{20}{c}}{\frac{c}{a}}& = &{ - \frac{b}{a}}\end{array}\]
So
\[ \Rightarrow \begin{array}{*{20}{c}}b& = &{ - c}\end{array}\]
Therefore, the value will be \[b = - c\]
Hence, the correct option is B.
Note: In this question, it is important to note that when the roots are decreased by the \[1\] and then the new roots of the equation will be \[\alpha - 1\]and \[\beta - 1\] the product of these roots will give the condition that will be satisfied by the sum and the product of the roots of the previous equation.
Formula Used:
Root’s sum can be determined by
\[\alpha + \beta = \frac{{ - b}}{a}\]
Root’s product can be determined by
\[\alpha \beta = \frac{c}{a}\]
Complete step by step solution:
We have been provided in the question that,
The equation formed by decreasing each root of \[a{x^2} + bx + c = 0\] by \[1\] is
\[2{x^2} + 8x + 2 = 0\]
Let us assume that the roots of the above equation are \[\alpha \]and \[\beta \]respectively.
We know the sum and the product of the roots of the above equation.
Therefore, we can write
\[\begin{array}{*{20}{c}}{ \Rightarrow \alpha + \beta }& = &{ - \frac{b}{a}}\end{array}\] -------- (1)
And
\[\begin{array}{*{20}{c}}{ \Rightarrow \alpha \beta }& = &{\frac{c}{a}}\end{array}\] ------------ (2)
Now we have given that when the roots of the above equation are decreased by \[1\], then the equation \[2\begin{array}{*{20}{c}}{{x^2} + 8x + 2}& = &0\end{array}\]is formed.
Therefore, the new roots will be \[\alpha - 1\]and \[\beta - 1\]respectively.
Now
The sum and the product of the roots of the formed equation will be
\[ \Rightarrow \begin{array}{*{20}{c}}{\left( {\alpha - 1} \right) + \left( {\beta - 1} \right)}& = &{ - \frac{8}{2}}\end{array}\]
\[\begin{array}{*{20}{c}}{ \Rightarrow \alpha + \beta }& = &{ - 2}\end{array}\]
And
\[ \Rightarrow \begin{array}{*{20}{c}}{\left( {\alpha - 1} \right)\left( {\beta - 1} \right)}& = &{\frac{2}{2}}\end{array}\]
\[ \Rightarrow \begin{array}{*{20}{c}}{\alpha \beta - \left( {\alpha + \beta } \right) + 1}& = &1\end{array}\]
\[ \Rightarrow \begin{array}{*{20}{c}}{\alpha \beta }& = &{\alpha + \beta }\end{array}\] -------- (3)
Now put the value of the value of the equation (1) and (2) in the equation (3). Therefore, we will get
\[ \Rightarrow \begin{array}{*{20}{c}}{\frac{c}{a}}& = &{ - \frac{b}{a}}\end{array}\]
So
\[ \Rightarrow \begin{array}{*{20}{c}}b& = &{ - c}\end{array}\]
Therefore, the value will be \[b = - c\]
Hence, the correct option is B.
Note: In this question, it is important to note that when the roots are decreased by the \[1\] and then the new roots of the equation will be \[\alpha - 1\]and \[\beta - 1\] the product of these roots will give the condition that will be satisfied by the sum and the product of the roots of the previous equation.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Degree of Dissociation and Its Formula With Solved Example for JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

Instantaneous Velocity - Formula based Examples for JEE

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

NCERT Solutions for Class 11 Maths Chapter 8 Sequences and Series
