• Nov 27th 2008, 09:35 AM
acc100jt
Find the range of values of $p$ for which $2x^{2}+4x+3>p$ is real for all values of $x$.

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

Thanks for those who help :)
• Nov 27th 2008, 05:38 PM
mr fantastic
Quote:

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

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 $2x^{2}+4x+3>p \Rightarrow 2x^2 + 4x + (3 - p) > 0$ so one way of solving the problem is to find the values of p such that the y-coordinate of the turning point of $y = 2x^2 + 4x + (3 - p)$ is greater than zero.
• Nov 27th 2008, 07:22 PM
Mathstud28
Quote:

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

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

Thanks for those who help :)

Let $z=3-p$

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

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

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

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

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

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

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

• Nov 27th 2008, 09:43 PM
mr fantastic
Quote:

Originally Posted by Mathstud28
Let $z=3-p$

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

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

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

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

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

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

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

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 28th 2008, 12:56 AM
CaptainBlack
Quote:

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

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 $p$ such that $2x^{2}+4x+3>p$ for all $x$.

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

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

or

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

has no real roots.

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

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

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

Which is again Mr Fantastics solution.

CB