CameraIcon

SearchIcon

MyQuestionIcon

483

You visited us 483 times! Enjoying our articles? Unlock Full Access!

Discriminant

Let α and ...

Question

Let $α$ and $β$ be nonzero real roots of the quadratic equation $x^{2} + a x + b = 0$ and $α + β, α - β, - α + β$ and $- α - β$ be the roots of the equation $x^{4} + a x^{3} + c x^{2} + d x + e = 0$ Then which of the following statement is false?

A

a = 0

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

B

c = 0

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

C

d = 0

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

D

e = 0

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

Open in App

Solution

The correct option is B $c = 0$
Given: $α, β$ are roots of the equation $x^{2} + a x + b = 0$ and $(α + β), (α - β), (β - α), (- (α + β))$ are roots of the equation $x^{4} + a x^{3} + c x^{2} + d x + e = 0$
To find which of the given statement is false
Sol: As $α, β$ are roots of the equation $x^{2} + a x + b = 0$
Hence $α + β = - a, α β = b . . . . . . . . . . (i)$
$(α + β), (α - β), (β - α), (- (α + β))$ are roots of the equation $x^{4} + a x^{3} + c x^{2} + d x + e = 0$
Hence, $(α + β) + (α - β) + (β - α) + (- (α + β)) = - a ⟹ a = 0... (i i)$
And,
$(α + β) (α - β) + (α + β) (β - α) + (α + β) (- (α + β)) + (α - β) (β - α) + (α - β) (- (α + β)) + (β - α) (- (α + β)) = c$
$⟹ (0) (α - β) + (0) (β - α) + (0) (- (0)) + (α - β) (β - α) + (α - β) (- (0)) + (β - α) (- (0)) = c$
[as $α + β = - a = 0$ from (i) and (ii)]
$⟹ (α - β) (β - α) = c$
$⟹ α β - α^{2} - β^{2} - α β = c ⟹ - (α^{2} + β^{2}) = c$
$⟹ - [(α + β)^{2} - 2 α β] = c$ [as $a^{2} + b^{2} = (a + b)^{2} - 2 a b$ ]
$⟹ - (0^{2} - 2 (b)) = c ⟹ c = 2 b$
And,
$(α + β) (α - β) (β - α) + (α + β) (α - β) (- (α + β)) + (α + β) (β - α) (- (α + β)) + (α - β) (β - α) (- (α + β)) = - d ⟹ (0) (α - β) (β - α) + (0) (α - β) (- (0)) + (0) (β - α) (- (0)) + (α - β) (β - α) (- (0)) = - d$
[as $α + β = - a = 0$ from (i) and (ii)]
$⟹ d = 0$
And
$(α + β) (α - β) (β - α) (- (α + β)) = e ⟹ e = 0$
[as $α + β = - a = 0$ from (i) and (ii)]

Suggest Corrections

thumbs-up

0

Similar questions

α, β

are the quadratic equation

x^{2} + a x + b = 0, (b \neq 0)

; then the quadratic equation whose roots are

α - \frac{1}{β}, β - \frac{1}{α}

Q. Statement I: lf

α, β

are the roots of

x^{2} - a x + b = 0

, then the equation whose roots are

\frac{α + β}{α}, \frac{α + β}{β}

b x^{2} - a^{2} x + a^{2} = 0

Statement II: lf

α, β

are the roots of

x^{2} - b x + c = 0

α + h, β + h

are the roots of

x^{2} + q x + r = 0

h = b - q

.
Which of the above statement(s) is(are) true.

α, β

are the roots of quadratic equation

a x^{3} + b x + c = 0

, then find the value of

\frac{α}{β} + \frac{β}{α}

α, β, γ

are the roots of

a x^{3} + b x^{2} + c x + d = 0

∣ ∣ ∣ ∣ ∣ \begin{matrix} α & β & γ β & γ & α γ & α & β \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0

α \neq β \neq γ

, then find the equation whose roots are

α + β - γ, γ + α - β, β + γ - α

α, β

be the roots of

x^{2} + b x + 1 = 0

. Then find the equation whose roots are

- (α + \frac{1}{β})

- (β + \frac{1}{α})

.

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Nature of Roots

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Standard X Mathematics

Join BYJU'S Learning Program

CrossIcon