
If the roots of the equation \[{x^2} + px + q = 0\;\] differ by$1$, then
(A) ${p^2} = 4q$
(B) ${p^2} = 4q + 1$
(C) ${p^2} = 4q - 1$
(D) None of these
Answer
163.2k+ views
Hint: In this question, we are given that the roots of the quadratic equation \[{x^2} + px + q = 0\;\] differ by$1$. We have to calculate the sum and the product of the roots, i.e., $\alpha + \beta = \dfrac{{ - B}}{A}$, $\alpha \beta = \dfrac{C}{A}$ where the equation is \[A{x^2} + Bx + C = 0\]. Then apply the algebraic identity ${\left( {\alpha + \beta } \right)^2} = {\left( {\alpha - \beta } \right)^2} + 4\alpha \beta$ and put in the required values.
Formula Used:
Quadratic equations in general: \[A{x^2} + Bx + C = 0\]
Let, the roots of the above quadratic equation be $\alpha$ and $\beta$
Therefore, the sum of roots is $\alpha + \beta = \dfrac{{ - B}}{A}$ and the product of roots is $\alpha \beta = \dfrac{C}{A}$
The algebraic identity is${\left( {\alpha + \beta } \right)^2} = {\left( {\alpha - \beta } \right)^2} + 4\alpha \beta$
Complete step by step Solution:
Given that, \[{x^2} + px + q = 0\;\], is a quadratic equation.
Let, the roots of the given equation be $\alpha$ and $\beta$
According to the question, the roots of the equation differ by$1$.
It implies that $\alpha - \beta = 1$ ------- (1)
Now, compare the equation \[{x^2} + px + q = 0\;\]with the general quadratic equation, i.e., \[A{x^2} + Bx + C = 0\].
We obtain$A = 1$,$B = p$, and$C = q$.
Using the formula for sum and the product of the roots, the sum of the roots,
\[\alpha + \beta = \dfrac{{ - B}}{A}\]
It implies that,
\[\alpha + \beta = - p\] --------(2)
And the product of the roots, $\alpha \beta = \dfrac{C}{A}$
$\Rightarrow \alpha \beta = q$ --------(3)
Now, applying the algebraic identity ${\left( {\alpha + \beta } \right)^2} = {\left( {\alpha - \beta } \right)^2} + 4\alpha \beta$
Substituting the required values in the above equation, (from equations (1), (2), and (3)),
It will be ${\left( { - p} \right)^2} = {\left( 1 \right)^2} + 4q$
On solving, we get ${p^2} = 4q+ 1$
Hence, the correct option is (B).
Note: The values of $x$ that fulfil a given quadratic equation \[A{x^2} + Bx + C = 0\] are known as its roots. They are, in other words, the values of the variables $\left( x \right)$ that satisfy the equation. The roots of a quadratic function are the $x -$coordinates of the function's $x -$intercepts. Because the degree of a quadratic equation is $2$, it can only have two roots.
Formula Used:
Quadratic equations in general: \[A{x^2} + Bx + C = 0\]
Let, the roots of the above quadratic equation be $\alpha$ and $\beta$
Therefore, the sum of roots is $\alpha + \beta = \dfrac{{ - B}}{A}$ and the product of roots is $\alpha \beta = \dfrac{C}{A}$
The algebraic identity is${\left( {\alpha + \beta } \right)^2} = {\left( {\alpha - \beta } \right)^2} + 4\alpha \beta$
Complete step by step Solution:
Given that, \[{x^2} + px + q = 0\;\], is a quadratic equation.
Let, the roots of the given equation be $\alpha$ and $\beta$
According to the question, the roots of the equation differ by$1$.
It implies that $\alpha - \beta = 1$ ------- (1)
Now, compare the equation \[{x^2} + px + q = 0\;\]with the general quadratic equation, i.e., \[A{x^2} + Bx + C = 0\].
We obtain$A = 1$,$B = p$, and$C = q$.
Using the formula for sum and the product of the roots, the sum of the roots,
\[\alpha + \beta = \dfrac{{ - B}}{A}\]
It implies that,
\[\alpha + \beta = - p\] --------(2)
And the product of the roots, $\alpha \beta = \dfrac{C}{A}$
$\Rightarrow \alpha \beta = q$ --------(3)
Now, applying the algebraic identity ${\left( {\alpha + \beta } \right)^2} = {\left( {\alpha - \beta } \right)^2} + 4\alpha \beta$
Substituting the required values in the above equation, (from equations (1), (2), and (3)),
It will be ${\left( { - p} \right)^2} = {\left( 1 \right)^2} + 4q$
On solving, we get ${p^2} = 4q+ 1$
Hence, the correct option is (B).
Note: The values of $x$ that fulfil a given quadratic equation \[A{x^2} + Bx + C = 0\] are known as its roots. They are, in other words, the values of the variables $\left( x \right)$ that satisfy the equation. The roots of a quadratic function are the $x -$coordinates of the function's $x -$intercepts. Because the degree of a quadratic equation is $2$, it can only have two roots.
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

IIT JEE Main Chemistry 2025: Syllabus, Important Chapters, Weightage

JEE Main Maths Question Paper PDF Download with Answer Key

JEE Main 2025 Session 2 City Intimation Slip Released - Download Link

Trending doubts
JEE Main Marks Vs Percentile Vs Rank 2025: Calculate Percentile Using Marks

JEE Mains 2025 Cutoff: Expected and Category-Wise Qualifying Marks for NITs, IIITs, and GFTIs

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

NIT Cutoff Percentile for 2025

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

JEE Main Syllabus 2025 (Updated)

Other Pages
NCERT Solutions for Class 10 Maths Chapter 13 Statistics

NCERT Solutions for Class 10 Maths Chapter 11 Areas Related To Circles

NCERT Solutions for Class 10 Maths Chapter 12 Surface Area and Volume

NCERT Solutions for Class 10 Maths Chapter 14 Probability

NCERT Solutions for Class 10 Maths In Hindi Chapter 15 Probability

Total MBBS Seats in India 2025: Government and Private Medical Colleges
