First slide

Theory of equations

Question

If $2 x^{3} + {ax}^{2} + bx + 4 = 0$ (a and b are positive real numbers) has three real roots.

NA

Question

The minimum value of a³ is

A

108
B

216
C

432
D

864

Solution

$∵ AM \geq GM$
$∴ \frac{(- α) + (- β) + (- γ)}{3} \geq {(- α) (- β) (- γ)}^{1 / 3}$
$\frac{\frac{a}{2}}{3} \geq (2)^{1 / 3}$
$\begin{aligned} a & \geq 6 (2)^{1 / 3} - - - - (i) \\ or a^{3} & \geq 432 \end{aligned}$
Hence, minimum value of a³ is 432.

Question

The minimum value of b³ is

A

432
B

864
C

1728
D

None of these

Solution

$∵ AM \geq GM$
$\begin{array}{lc} ∴ & \frac{(- α) (- β) + (- β) (- γ) + (- γ) (- α)}{3} \\ \Rightarrow & \geq {(- α) (- β) (- β) (- γ) (- γ) (- α)}^{1 / 3} \\ \Rightarrow & \frac{b / 2}{3} \geq (4)^{1 / 3} - - - - (ii) \\ \Rightarrow & b \geq 6 (4)^{1 / 3} \end{array}$
or $b^{3} \geq 864$
Hence, minimum value of b³ is 864

Question

The minimum value of (a + b)³ is

A

1728
B

3456
C

6912
D

864

Solution

From Eqs. (i) and (iü), we get
$ab \geq 6 (2)^{1 / 3} \cdot 6 (4)^{1 / 3}$
$\begin{array}{ll} \Rightarrow ab & \geq 36 \times 2 \\ ∵ \frac{a + b}{2} & \geq \sqrt{ab} \geq 6 \sqrt{2} \Rightarrow \frac{a + b}{2} \geq 6 \sqrt{2} \\ ∴ a + b & \geq 12 \sqrt{2} \\ or (a + b)^{3} & \geq 3456 \sqrt{2} \end{array}$
Hence, minimum value of (a + b)³ is $3456 \sqrt{2}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App