Let k be a real number such that k- 0- If - and - are non zero complex numbers satisfying -2k and -2-2-4k2-2k- then a quadratic equation having - and - as its roots is equal to

The correct option is B x^2 4kx+4k=0 Given α+β= 2k and α^2+β^2=4k^2 2k , α^2+β^2=(α+β) 2αβ ⇒αβ=k Now, sum of roots =α+β/α+α+β/β ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Complex Numbers

Let k be a ...

Question

Let $k$ be a real number such that $k \neq 0$ . If $α$ and $β$ are non zero complex numbers satisfying $α + β = - 2 k$ and $α^{2} + β^{2} = 4 k^{2} - 2 k$ , then a quadratic equation having $\frac{α + β}{α}$ and $\frac{α + β}{β}$ as its roots is equal to

4 x^{2} - 4 k x + k = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x^{2} - 4 k x + 4 k = 0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

4 k x^{2} - 4 x + k = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4 k x^{2} - 4 k x + 1 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $x^{2} - 4 k x + 4 k = 0$
Given $α + β = - 2 k$ and $α^{2} + β^{2} = 4 k^{2} - 2 k$ , $α^{2} + β^{2} = (α + β) - 2 α β$
$\Rightarrow α β = k$
Now, sum of roots $= \frac{α + β}{α} + \frac{α + β}{β}$ $= - 2 k (\frac{1}{α} + \frac{1}{β})$
$= 4 k$
Product of roots $= (\frac{α + β}{α}) (\frac{α + β}{β})$
$= \frac{(α + β)^{2}}{α β}$
$= 4 k$
So, the quadratic equation is $x^{2} - 4 k x + 4 k = 0$ .

Suggest Corrections

Similar questions

Q. Let

p

and

q

be real numbers such that

p \neq 0, p^{3} \neq q

and

p^{3} \neq - 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

\frac{α}{β}

and

\frac{β}{α}

as its roots is

Q. Let

p

and

q

be real numbers such that

p \neq 0, p^{3} \neq q

and

p^{3} \neq - q

. If

α

and

β

are nonzero complex numbers satisfying

α + β = - p

and

α^{3} + β^{3} = - q

, then a quadratic equation having

\frac{α}{β}

and

\frac{β}{α}

as its roots is

Q. Let p and q be real numbers such that

p \neq 0, p^{3} \neq q

and

p^{3} \neq - 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

\frac{α}{β}

and

\frac{β}{α}

as its roots is

Q. If

α

and

β

are the roots of the equation.

3 x^{2} - 4 x + 1 = 0

, from a quadratic equation whose roots are

\frac{α^{2}}{β}

and

\frac{β^{2}}{α}

Q. If

α

and

β

are the roots of the equation

x^{2} - 4 \sqrt{2} k x + 2 e^{4 l n k} - 1 = 0

for some

k

and

α^{2} + β^{2} = 66

, then

α^{2} + β^{2} + α + β

is equal to

Join BYJU'S Learning Program

Let k be a real number such that k≠0. If α and β are non zero complex numbers satisfying α+β=−2k and α2+β2=4k2−2k, then a quadratic equation having α+βα and α+ββ as its roots is equal to

The correct option is B x2−4kx+4k=0Given α+β=−2k and α2+β2=4k2−2k,α2+β2=(α+β)−2αβ⇒αβ=kNow, sum of roots =α+βα+α+ββ =−2k(1α+1β) =4kProduct of roots =(α+βα)(α+ββ) =(α+β)2αβ =4kSo, the quadratic equation is x2−4kx+4k=0.

Let $k$ be a real number such that $k \neq 0$ . If $α$ and $β$ are non zero complex numbers satisfying $α + β = - 2 k$ and $α^{2} + β^{2} = 4 k^{2} - 2 k$ , then a quadratic equation having $\frac{α + β}{α}$ and $\frac{α + β}{β}$ as its roots is equal to