quadratic inequality

**acc100jt** · November 27th 2008, 08:35 AM

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

**mr fantastic** · November 27th 2008, 04:38 PM

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.

**Mathstud28** · November 27th 2008, 06:22 PM

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

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

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

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

And that is your answer.

**mr fantastic** · November 27th 2008, 08:43 PM

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

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

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

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

And that is your answer.

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.

**CaptainBlack** · November 27th 2008, 11:56 PM

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

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