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

Byju's Answer

Standard X

Mathematics

Nature of Roots

If the roots ...

Question

If the roots of the equation $b x^{2} + c x + a = 0$ be imaginary, then for all real values of $x$ , the expression $3 b^{2} x^{2} + 6 b c x + 2 c^{2}$ is

greater than

- 2 a b

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

less than

- 2 a b

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

greater than

- 4 a b

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

less than

- 4 a b

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

Open in App

Solution

The correct option is D greater than $- 4 a b$
As, $b x^{2} +$ $c x$ $+ a = 0$ has imaginary roots .

So, $c^{2} < 4 a b$

Now,

$3 b^{2} x^{2} + 6 b c x + 2 c^{2}$

$= 3 (b x + c)^{2} - c^{2} \geq - c^{2} \geq - 4 a b$

Hence, option 'C' is correct .

Suggest Corrections

Similar questions

If the roots of the equation $b x^{2} + c x + a = 0$ be nonreal, then for all real values of x, the expression $3 b^{2} x^{2} + 6 b c x + 2 c^{2}$ is

Q. (a) If the roots

α, β

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

be real, then establish between the coefficients under the following conditions:
(i) Roots are equal and opposite.
(ii) Roots are of opposite signs.
(iii) Roots are both

- i v e

(iv) Roots are both

+ i v e

.
(b) Let

a > 0, b > 0

, then both roots of the equation

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

(i) are real and negative
(ii) have negative real parts.
(iii) none of these.
(c) If the roots of the equation

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

be imaginary. then for all real values of x, the expression

3 b^{2} x^{2} + 6 b c x + 2 c^{2}

is
(a) less than

- 4 a b

(b) greater then

4 a b

4 a b

(d) greater than

- 4 a b

Q. Let the equation

x^{2}

+ 2(a - 1)x + a + 5 = 0, where 'a' is a parameter, match the real value of 'a' so that the given equation has :-

Column - I	Column - II
(A) Imaginary roots	(P) $(- \infty, - \frac{8}{7})$
(B) One root less than 3 other root is greater than 3	(Q) (-1, 4)
(C) One root less than 1 & other root is greater than 3	(R) $(- \infty, - \frac{4}{3})$

Q. For what real values of

k

, other than

k = 0

, does the equation

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

have real roots? (The symbol

x \geq a

means that

x

can take on all values greater than

a

and the value

a

itself;

x \leq a

has the corresponding meaning with "less than")

Q. For values of

x

less than

1

but greater than

- 4

, the expression

\frac{x^{2} - 2 x + 2}{2 x - 2}

has

Join BYJU'S Learning Program

If the roots of the equation bx2+cx+a=0 be imaginary, then for all real values of x, the expression 3b2x2+6bcx+2c2 is

The correct option is D greater than −4abAs, bx2+ cx +a=0 has imaginary roots .So, c2<4abNow, 3b2x2+6bcx+2c2=3(bx+c)2−c2≥−c2≥−4ab Hence, option 'C' is correct .

If the roots of the equation $b x^{2} + c x + a = 0$ be imaginary, then for all real values of $x$ , the expression $3 b^{2} x^{2} + 6 b c x + 2 c^{2}$ is

The correct option is D greater than $- 4 a b$
As, $b x^{2} +$ $c x$ $+ a = 0$ has imaginary roots .

So, $c^{2} < 4 a b$

Now,

$3 b^{2} x^{2} + 6 b c x + 2 c^{2}$

$= 3 (b x + c)^{2} - c^{2} \geq - c^{2} \geq - 4 a b$

Hence, option 'C' is correct .