Let p and q be real numbers such that p ≠ 0, p^3 ≠ q and p^3 ≠-q . If α and β are non-zero complex numbers satisfying α+β=-p and α^3+β^3=q , then a quadratic equation having (α)/(β) and (β)/(α) as its roots is

Correct answer: B. (p^3+q) x^2-(p^3-2 q) x +(p^3+q)=0 Sum of roots =(α^2+β^2)/(α β) and product =1 Given, α+β=-p and α^3+β^3=q aligned & (α+β)(α^2-α β+β^2)=q & α^2+β^2-α β=(-q)/(p) aligned and (α+β)^2=p^2 α^ 2+ β^ 2+2 α β= p^2 From Eqs. (i) and (ii), we get α^ 2+ β^ 2= (p^3-2 q)/(3 p) and α β=(p^3+q)/(3 p) Required equation is gathered x²-((p^3-2 q) x)/((p^3+q))+1=0 (p^3+q) x²-(p^3-2 q) x+(p^3+q)=0 gathered

JEE Advanced 2010Paper 1

Exam: JEE Advanced
Year: 2010
Paper: Paper 1
Subject: Mathematics
Chapter: Quadratic Equation

JEE Advanced 2010 · Paper 1 · Mathematics · Quadratic Equation

Let p and q be real numbers such that p ≠ 0, p^3 ≠ q and p^3 ≠-q ....

Let

p

and

q

be real numbers such that

p \neq 0, p^3 \neq q

and

p^3 \neq-q

. If

\alpha

and

\beta

are non-zero complex numbers satisfying

\alpha+\beta=-p

and

\alpha^3+\beta^3=q

, then

a

quadratic equation having

\frac{\alpha}{\beta}

and

\frac{\beta}{\alpha}

as its roots is

Options

A $\left(p^3+q\right) x^2-\left(p^3+2 q\right) x$ $+\left(p^3+q\right)=0$
B $\left(p^3+q\right) x^2-\left(p^3-2 q\right) x$ $+\left(p^3+q\right)=0$
C $\left(p^3-q\right) x^2-\left(5 p^3-2 q\right) x$ $+\left(p^3-q\right)=0$
D $\left(p^3-q\right) x^2-\left(5 p^3+2 q\right) x$ $+\left(p^3-q\right)=0$

Correct answer

B. $\left(p^3+q\right) x^2-\left(p^3-2 q\right) x$ $+\left(p^3+q\right)=0$

Step-by-step solution

Sum of roots

=\frac{\alpha^2+\beta^2}{\alpha \beta}

and product

=1

Given,

\alpha+\beta=-p

and

\alpha^3+\beta^3=q

\begin{aligned}& \Rightarrow(\alpha+\beta)\left(\alpha^2-\alpha \beta+\beta^2\right)=q \\ & \therefore \quad \alpha^2+\beta^2-\alpha \beta=\frac{-q}{p} \end{aligned}

and

\quad(\alpha+\beta)^2=p^2

\Rightarrow \quad

\alpha^

\beta^

2+2

\alpha \beta=

p^2 From Eqs. (i) and (ii), we get

\alpha^

\beta^

\frac{p^3-2 q}{3 p}

and

\alpha \beta=\frac{p^3+q}{3 p}\therefore

Required equation is \begin{gathered}x²-\frac{\left(p^3-2 q\right) x}{\left(p^3+q\right)}+1=0 \\ \Rightarrow\left(p^3+q\right) x²-\left(p^3-2 q\right) x+\left(p^3+q\right)=0 \end{gathered}

Practice Quadratic Equation on Quantrex Academy →

More from Quadratic Equation