Let - - - be the roots of ax2-bx -c - 0-a- 0 and - 1- - - be the roots of a1x2- b1x - c1 - 0-a1 - 0- Then the quadratic equation whose roots are - - 1 is

The correct option is D x^2 ( ba +b_1a_1 )+ x + 1 ( bc +b_1c_1 )=0α , β are the roots of ax^2+bx +c = 0 ⇒α + β = ba, αβ = ca α _1, β are the roots of a_1x^2 ...

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

Byju's Answer

Standard XIII

Mathematics

Relation between Roots and Coefficients for Quadratic

Let α , β be ...

Question

Let $α, β$ be the roots of $a x^{2} + b x + c = 0, a \neq 0$ and $α_{1}, - β$ be the roots of $a_{1} x^{2} + b_{1} x + c_{1} = 0, a_{1} \neq 0$ . Then the quadratic equation whose roots are $α, α_{1}$ is

\frac{x^{2}}{(\frac{b}{a} + \frac{b_{1}}{a_{1}})} - x + \frac{1}{(\frac{b}{c} + \frac{b_{1}}{c_{1}})} = 0

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

\frac{x^{2}}{(\frac{b}{a} + \frac{b_{1}}{a_{1}})} + 2 x + \frac{1}{(\frac{b}{c} - \frac{b_{1}}{c_{1}})} = 0

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

\frac{2 x^{2}}{(\frac{b}{a} + \frac{b_{1}}{a_{1}})} + x + \frac{1}{(\frac{b}{c} + \frac{b_{1}}{c_{1}})} = 0

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

\frac{x^{2}}{(\frac{b}{a} + \frac{b_{1}}{a_{1}})} + x + \frac{1}{(\frac{b}{c} + \frac{b_{1}}{c_{1}})} = 0

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

Open in App

Solution

The correct option is D $\frac{x^{2}}{(\frac{b}{a} + \frac{b_{1}}{a_{1}})} + x + \frac{1}{(\frac{b}{c} + \frac{b_{1}}{c_{1}})} = 0$
$α, β$ are the roots of $a x^{2} + b x + c = 0$
$\Rightarrow α + β = - \frac{b}{a}, α β = \frac{c}{a}$
$α_{1}, - β$ are the roots of $a_{1} x^{2} + b_{1} x + c_{1} = 0$
$\Rightarrow α_{1} - β = - \frac{b_{1}}{a_{1}}, - α_{1} β = \frac{c_{1}}{a_{1}}$

Now, $(α + β) + (α_{1} - β) = - \frac{b}{a} - \frac{b_{1}}{a_{1}} = α + α_{1}$
$\frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β} = \frac{- b / a}{c / a} = - \frac{b}{c} \dots (1)$
$\frac{1}{α_{1}} - \frac{1}{β} = \frac{β - α_{1}}{α_{1} β} = \frac{- b_{1}}{c_{1}} \dots (2)$

From $(1)$ and $(2)$ ,
$\frac{1}{α} + \frac{1}{α_{1}} = - (\frac{b}{c} + \frac{b_{1}}{c_{1}}) = \frac{α + α_{1}}{α α_{1}}$
$∴ α α_{1} = \frac{\frac{b}{a} + \frac{b_{1}}{a_{1}}}{\frac{b}{c} + \frac{b_{1}}{c_{1}}}$

The equation whose roots are $α, α_{1}$ is
$x^{2} - (α + α_{1}) x + α α_{1} = 0$
$\Rightarrow x^{2} + (\frac{b}{a} + \frac{b_{1}}{a_{1}}) x + \frac{\frac{b}{a} + \frac{b_{1}}{a_{1}}}{\frac{b}{c} + \frac{b_{1}}{c_{1}}} = 0$
$\Rightarrow \frac{x^{2}}{(\frac{b}{a} + \frac{b_{1}}{a_{1}})} + x + \frac{1}{(\frac{b}{c} + \frac{b_{1}}{c_{1}})} = 0$

Suggest Corrections

Similar questions

Q. If

α, β

are the roots of

a x^{2} + b x + c = 0, α_{1}, - β

are the roots of

a_{1} x^{2} + b_{1} x + c_{1} = 0

, show that

α, α_{1}

are the roots of

\frac{x^{2}}{\frac{b}{a} + \frac{b_{1}}{a_{1}}} + x + \frac{1}{\frac{b}{c} + \frac{b_{1}}{c_{1}}} = 0

Q. Assertion :If

\frac{a}{a_{1}}, \frac{b}{b_{1}}, \frac{c}{c_{1}}

are in

A . P

then

a_{1}, b_{1}, c_{1}

are in

G . P

Reason: If

a x^{2} + b x + c

and

a_{1} x^{2} + b_{1} x^{2} + c_{1} = 0

have a common root &

\frac{a}{a_{1}}, \frac{b}{b_{1}}, \frac{c}{c_{1}}

are in

A . P

then

a_{1}, b_{1}, c_{1}

are in

G . P

Q. Let

a, b, c, a_{1}, b_{1}, c_{1}, \in R

and

a x^{2} + b x + c > 0 \forall x \in R

and

a_{1} x^{2} + b_{1} x + c_{1} > 0 \forall x \in R .

Then

Q. Let

a, b, c, a_{1}, b_{1}, c_{1}, \in R

and

a x^{2} + b x + c > 0 \forall x \in R

and

a_{1} x^{2} + b_{1} x + c_{1} > 0 \forall x \in R .

Then

Q. Prove that

(b - c) cot \frac{1}{2} A + (c - a) cot \frac{1}{2} B + (a - b) cot \frac{1}{2} C = 0

Join BYJU'S Learning Program

Let α,β be the roots of ax2+bx+c=0,a≠0 and α1,−β be the roots of a1x2+b1x+c1=0,a1≠0. Then the quadratic equation whose roots are α,α1 is

Let $α, β$ be the roots of $a x^{2} + b x + c = 0, a \neq 0$ and $α_{1}, - β$ be the roots of $a_{1} x^{2} + b_{1} x + c_{1} = 0, a_{1} \neq 0$ . Then the quadratic equation whose roots are $α, α_{1}$ is