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

Relations between Roots and Coefficients : Higher Order Equations

Assertion :Th...

Question

Assertion :The number of minimum possible complex roots of the equation $x^{6} - 3 x^{5} + 4 x^{3} + 3 x^{2} + 4 = 0$ is $2$ Reason: The equation $x^{6} - 3 x^{5} + 4 x^{3} + 3 x^{2} + 4 = 0$ has maximum four real roots.

Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

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

Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

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

Assertion is correct but Reason is incorrect

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

Both Assertion and Reason are incorrect

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

Open in App

Solution

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
Since $f (x)$ has two changes in sign & $f (- x)$ has too, two changes in sign. Hence the maximum number of real roots is 4 and hence the equation has at least (minimum) two complex roots.

Suggest Corrections

Similar questions

x^{6} - 3 x^{5} + 4 x^{3} + 3 x^{2} + 4 = 0

f (x) = x^{6} - 3 x^{5} + 4 x^{3} + 3 x^{2} + 4 \Rightarrow f (x)

Assertion: All the points (1, 0), (-1, 0), (2, 0) and (5, 0) lie on the x – axis.

Reason: Equation of the x – axis is y = 0
Now, choose the correct option.

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

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

(b^{2} - 4 a c) < 0 .

y = A sin (k x - ω t)

x = 0

y -

x -

H_{2}

H e

(1 + \frac{P b}{R T})

H_{2}

H e

Join BYJU'S Learning Program