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

Derivative from First Principle

If the roots ...

Question

If the roots of the equation $a x^{2} + x + b = 0$ be real and distinct, then the roots of the equation $x^{2} - 4 \sqrt{a b} x + 1 = 0$ will be-

rational

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

real

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

imaginary

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

irrational

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

Open in App

Solution

The correct option is C imaginary
${a x}^{2} + x + b = 0 r o o t s a r e r e a l a n d d i s t i n c t ∴ D > 0 1 - 4 a b > 0$
$D_{1} = 4 a b - 1 = - D < 0 ∴ D_{1} < 0 s o r o o t s a r e i m a g i n a r y$

Suggest Corrections

Similar questions

f (x) {(f^{''} (x))}^{2} + f (x) f^{'} (x) f^{'''} (x) + {(f^{'} (x))}^{2} f^{''} (x) = 0

f (x) = 0

f (x) = 0

f^{'} (x) = 0

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

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

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

a, b, c

m

n

m^{2} > n > 0

{a x}^{2} + 2 m b x + n c = 0

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

Join BYJU'S Learning Program