• Nov 27th 2008, 08:35 AM
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 :)
• Nov 27th 2008, 04:38 PM
mr fantastic
Quote:

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.
• Nov 27th 2008, 06:22 PM
Mathstud28
Quote:

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 $\displaystyle 3\leqslant{p}$

• Nov 27th 2008, 08:43 PM
mr fantastic
Quote:

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.
• Nov 27th 2008, 11:56 PM
CaptainBlack
Quote:

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