
If ${x^2} - 3x + 2$ be a factor of ${x^4} - p{x^2} + q$ , then $(p,q) = $
A. $(3,4)$
B. $(4,5)$
C. $(4,3)$
B. $(5,4)$
Answer
164.7k+ views
Hint: We are provided here two expressions. One expression is quadratic and a factor of the other expression. So, we can find the roots of the first expression using the Sridharacharya formula. And as we know, if one expression is a factor of another then, the roots of the equation will also satisfy the other equation. We have to use this concept only and further we will get two simultaneous equations. On solving, we will get the final result.
Formula Used: The roots of a quadratic equation are found using the quadratic formula.
$\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Complete step-by-step solution:
We have been given two expressions as follows:
${x^2} - 3x + 2$ --(1.1)
And
${x^4} - p{x^2} + q$ --(1.2)
As given, the equation (1.1) is a factor of the equation (1.2). So, first we will find the roots of the equation (1.1).
To find the roots, compare equation (1.1) with the standard quadratic equation $a{x^2} + bx + c = 0$ .
We get $a = 1$ , $b = - 3$ and $c = 2$ .
Substituting these values in the quadratic formula $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ to get the value for $x$
$\therefore $ $x = \dfrac{{ - ( - 3) \pm \sqrt {{{( - 3)}^2} - 4 \times 1 \times 2} }}{{2 \times 1}}$ .
$ \Rightarrow x = \dfrac{{3 \pm \sqrt 1 }}{2}$
So, we have
$ \Rightarrow x = 2,1$
Now, we substitute these values of $x$ in the equation (1.2), we will get following two expressions:
${(2)^4} - {(2)^2}p + q$ and $p - q - 1$ .
To get the value of the unknowns we equalize these to zero, to express them as equations.
$4p - q - 16 = 0$ and $p - q - 1 = 0$
Solving these equations simultaneously, we get the result that
$(p,q) = (5,4)$ .
Hence, the correct option is D.
Note: It should be kept in mind that the first equation is a factor of the other equation. So, we will only find the roots of the first quadratics. If we find the roots of the second expression then, it will not satisfy the first expression as the second expression is not a factor of the first one. Taking that under consideration, we have to satisfy the second expression only.
Formula Used: The roots of a quadratic equation are found using the quadratic formula.
$\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Complete step-by-step solution:
We have been given two expressions as follows:
${x^2} - 3x + 2$ --(1.1)
And
${x^4} - p{x^2} + q$ --(1.2)
As given, the equation (1.1) is a factor of the equation (1.2). So, first we will find the roots of the equation (1.1).
To find the roots, compare equation (1.1) with the standard quadratic equation $a{x^2} + bx + c = 0$ .
We get $a = 1$ , $b = - 3$ and $c = 2$ .
Substituting these values in the quadratic formula $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ to get the value for $x$
$\therefore $ $x = \dfrac{{ - ( - 3) \pm \sqrt {{{( - 3)}^2} - 4 \times 1 \times 2} }}{{2 \times 1}}$ .
$ \Rightarrow x = \dfrac{{3 \pm \sqrt 1 }}{2}$
So, we have
$ \Rightarrow x = 2,1$
Now, we substitute these values of $x$ in the equation (1.2), we will get following two expressions:
${(2)^4} - {(2)^2}p + q$ and $p - q - 1$ .
To get the value of the unknowns we equalize these to zero, to express them as equations.
$4p - q - 16 = 0$ and $p - q - 1 = 0$
Solving these equations simultaneously, we get the result that
$(p,q) = (5,4)$ .
Hence, the correct option is D.
Note: It should be kept in mind that the first equation is a factor of the other equation. So, we will only find the roots of the first quadratics. If we find the roots of the second expression then, it will not satisfy the first expression as the second expression is not a factor of the first one. Taking that under consideration, we have to satisfy the second expression only.
Recently Updated Pages
Trigonometry Formulas: Complete List, Table, and Quick Revision

Difference Between Distance and Displacement: JEE Main 2024

IIT Full Form

Uniform Acceleration - Definition, Equation, Examples, and FAQs

Difference Between Metals and Non-Metals: JEE Main 2024

Newton’s Laws of Motion – Definition, Principles, and Examples

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

NIT Cutoff Percentile for 2025

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

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

JEE Main 2025 CutOff for NIT - Predicted Ranks and Scores

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
