Question

Find the condition for 2x^2 + 3kx + 4c to be positive for all real values of x.​

Answer 1

Answer:

Show that x2+2px+2p2 is positive for all real values of x.

I've worked it out like so:

Discriminant = (2p)2−(4×1×(2p2))=4p2−8p2

I realise that the discriminant must be ≤0

No matter the value of p, 4p2−8p2 will always be ≤0.

Also, by completing the square:

x2+2px+2p2=(x+p)2+p2

Again, the p2 value on the right will always be positive.

Therefore, no matter the value of p, the parabola will be positive for all values of y.

Am I correct?

I feel that there could be a more mathematical way of expressing this.

Answer 2

Given: An expression $2 {x}^{2} + 3kx + 4c$ is given

To find: Condition for this expression to be positive for all values of x

Explanation: A quadratic expression

$a {x}^{2} + bx + r$

is always positive in nature when a>0 and d<0 where

$d = \sqrt{ {b}^{2} - 4ar}$

If d is greater than or equal to 0, then the expression has 1 or more roots which will make the value of the expression 0 at some point. Therefore, d<0.

In this equation, a=2 , b= 3k and r= 4c

Here, a>0 and

$d = \sqrt{ {(3k)}^{2} - 4 \times 2 \times 4c}$

$d = \sqrt{9 {k}^{2} - 32c}$

For this expression to be positive for all values of x, the value of $\sqrt{ 9 {k}^{2} - 32c}$ should be less than 0.