New chat PYQ Library Doubts Rank report

Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →

Quadratic Equation and Inequalities question

2009 · Shift 0 · Q44

Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.

PYQ Library
JEE Main
Mathematics
Quadratic Equation and Inequalities
2009 · Shift 0 · Q44

Quadratic Equation and Inequalities question

2009 · Shift 0 · Q44

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1

If the roots of the equation

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

imaginary, then for all real values of

x

, the expression

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

is :

A
less than $4ab$
B
greater than $-4ab$
C
less than $-4ab$
D
greater than $4ab$

View written solutionFree

Correct answer: B

Given quadratic:

$bx^2+cx+a=0$

Its roots are imaginary, so its discriminant is negative:

$c^2-4ab<0$

Hence,

$c^2<4ab$

We need to study the expression

$E=3b^2x^2+6bcx+2c^2$

Rewrite it by completing the square:

$E=3(b^2x^2+2bcx)+2c^2$

Now,

$b^2x^2+2bcx=(bx+c)^2-c^2$

So,

$E=3[(bx+c)^2-c^2]+2c^2$

$E=3(bx+c)^2-c^2$

Since $(bx+c)^2\ge 0$ for all real $x$ ,

$3(bx+c)^2\ge 0$

Therefore,

$E=3(bx+c)^2-c^2\ge -c^2$

Also, equality occurs when

$bx+c=0$

From the imaginary roots condition,

$c^2<4ab$

Multiplying by $-1$ reverses the inequality:

$-c^2>-4ab$

Since

$E\ge -c^2$

we get

$E>-4ab$

for all real $x$ .

Now check options:

A: $E<4ab$ — not always true.
B: $E>-4ab$ — true for all real $x$ .
C: $E<-4ab$ — false.
D: $E>4ab$ — not always true.

Therefore the correct option is:

$\boxed{\text{B}}$

More from Quadratic Equation and Inequalities

The quadratic equations $x^{2} - 6 x + a = 0$ and $x^{2} - c x + 6 = 0$ have one root in common. The other roots of the first and second equations are integers in the ratio 4 : 3. Then the common root is2008 · MCQ
STATEMENT - 1 : For every natural number $n \geq 2, \frac{1}{1} + \frac{1}{2} + ........ + \frac{1}{n} > n .$ STATEMENT - 2 : For every natural number $n \geq 2,$ , $n (n + 1) < n + 1.$ 2008 · MCQ
If the difference between the roots of the equation $x^{2} + a x + 1 = 0$ is less than $5,$ then the set of possible values of $a$ is2007 · MCQ
If the roots of the quadratic equation $x^{2} + p x + q = 0$ are $tan 3 0^{\circ}$ and $tan 1 5^{\circ}$ , respectively, then the value of $2 + q - p$ is2006 · MCQ
All the values of $m$ for which both roots of the equation $x^{2} - 2 m x + m^{2} - 1 = 0$ are greater than $- 2$ but less then 4, lie in the interval2006 · MCQ
If $x$ is real, the maximum value of $\frac{3 x ^{2} + 9 x + 17}{3 x ^{2} + 9 x + 7}$ is2006 · MCQ
The value of $a$ for which the sum of the squares of the roots of the equation $x^{2} - (a - 2) x - a - 1 = 0$ assume the least value is2005 · MCQ
If the roots of the equation $x^{2} - b x + c = 0$ be two consecutive integers, then $b^{2} - 4 c$ equals2005 · MCQ