Find the range of values of $\displaystyle p$ for which $\displaystyle 2x^{2}+4x+3>p$ is real for all values of $\displaystyle x$.

What's the meaning of the "is real" here??
Does it mean "is true"?

Thanks for those who help

2. Originally Posted by acc100jt
Find the range of values of $\displaystyle p$ for which $\displaystyle 2x^{2}+4x+3>p$ is real for all values of $\displaystyle x$.

What's the meaning of the "is real" here??
Does it mean "is true"?

Thanks for those who help
It means that the solution to the inequality is all values of x.

Note that $\displaystyle 2x^{2}+4x+3>p \Rightarrow 2x^2 + 4x + (3 - p) > 0$ so one way of solving the problem is to find the values of p such that the y-coordinate of the turning point of $\displaystyle y = 2x^2 + 4x + (3 - p)$ is greater than zero.

3. Originally Posted by acc100jt
Find the range of values of $\displaystyle p$ for which $\displaystyle 2x^{2}+4x+3>p$ is real for all values of $\displaystyle x$.

What's the meaning of the "is real" here??
Does it mean "is true"?

Thanks for those who help
Let $\displaystyle z=3-p$

Then we must find all values of x such that $\displaystyle 2x^2+4x+z>0$

Now solving $\displaystyle 2x^24x+z=0$ we get $\displaystyle x=\frac{-(\sqrt{-2(x-2)}+2)}{2}$ and $\displaystyle x=\frac{\sqrt{-2(z-2)}-2}{2}$

Now since this is an upward facing parabola we want all the values to the "left" and "right" of the two zeros, so we get

$\displaystyle x<\frac{-(\sqrt{-2(x-2)}+2)}{2}$ and $\displaystyle \frac{\sqrt{-2(z-2)}-2}{2}<x$

Now sub back $\displaystyle z=3-p$ to get

$\displaystyle x<\frac{-(\sqrt{2p-6}+2)}{2}$ and $\displaystyle \frac{\sqrt{2p-6}-2}{2}<x$

So for this value to be real we need to have that $\displaystyle 0\leqslant{2p-6}$ or in other words $\displaystyle 3\leqslant{p}$

4. Originally Posted by Mathstud28
Let $\displaystyle z=3-p$

Then we must find all values of x such that $\displaystyle 2x^2+4x+z>0$

Now solving $\displaystyle 2x^24x+z=0$ we get $\displaystyle x=\frac{-(\sqrt{-2(x-2)}+2)}{2}$ and $\displaystyle x=\frac{\sqrt{-2(z-2)}-2}{2}$

Now since this is an upward facing parabola we want all the values to the "left" and "right" of the two zeros, so we get

$\displaystyle x<\frac{-(\sqrt{-2(x-2)}+2)}{2}$ and $\displaystyle \frac{\sqrt{-2(z-2)}-2}{2}<x$

Now sub back $\displaystyle z=3-p$ to get

$\displaystyle x<\frac{-(\sqrt{2p-6}+2)}{2}$ and $\displaystyle \frac{\sqrt{2p-6}-2}{2}<x$

So for this value to be real we need to have that $\displaystyle 0\leqslant{2p-6}$ or in other words $\displaystyle 3\leqslant{p}$

My understanding of the question is that the inequality has to be satisfied for all real values of x. If that's the case, the solution will be as I suggested in my first post. The answer will be p < 1.

5. Originally Posted by acc100jt
Find the range of values of $\displaystyle p$ for which $\displaystyle 2x^{2}+4x+3>p$ is real for all values of $\displaystyle x$.

What's the meaning of the "is real" here??
Does it mean "is true"?

Thanks for those who help
Lets assume you are asked to find those values of $\displaystyle p$ such that $\displaystyle 2x^{2}+4x+3>p$ for all $\displaystyle x$.

Well as we have a parabola opening upwards this means that we seek values of $\displaystyle p$ such that:

$\displaystyle 2x^{2}+4x+3=p$

or

$\displaystyle 2x^{2}+4x+3-p=0$

has no real roots.

Using the quadratic formula, we find the roots to be:

$\displaystyle x=\frac{-4\pm \sqrt{16-8(3-p)}}{4}$

and for there to be no real roots requires that $\displaystyle 16-8(3-p)<0$ or $\displaystyle p<1$.

Which is again Mr Fantastics solution.

CB