1. ## Function

Hi there, I have a problem that I can not seem to figure out. It is asking on the condition that $a >0$ to prove that the quadratic expression $ax^2 +bx +c$ is positve for all real values of $x$ when $b^2 < 4ac$. then it saids based on that to find the range of values of p for which the quadratic function of $x$

$
f(x) \equiv 4x^2 + 4px - (3p^2 +4p -3)
$

is positive for all real values of $x$ illustrate your result by making sketch graphs of $f(x)$ for each of the cases $p=0$ and $p=1$

Can some one out there help me to figure out this problem/

2. First question : First, you know that $a> 0$. That means that if you sketch the graph for any permitted value of $p$, it will be a curved parabola open to the top. Therefore, there are only two possible situations : either it never touches the x-axis, therefore there are no real solutions and the equation is always positive, either it cuts the x-axis and there are solutions. But the equation cannot be all negative. Now, if $b^2 < 4ac$, that means that $b^2 - 4ac < 0$. Basically, it means that the discriminant is negative, thus the equation admits no real solution and therefore is positive for all real $x$ (you got to rewrite it properly though, I gave you the idea).

Second question : you need to find all values of $p$ for which the given quadratic has only positive values (that is, has no real solution). Think of your equation like this :

$ax^2 + bx + c$

Where :

$a = 4$
$b = 4p$
$c = -(3p^2 + 4p - 3)$

Get the discriminant of this equation :

$\Delta = b^2 - 4ac$

$\Delta = (4p)^2 - 4 \times 4 \times [-(3p^2 + 4p - 3)]$

Now, what is the characteristic feature of a quadratic equation with no real solution (i.e. all values are positive, in our case) ? The discriminant is negative. Therefore you have :

$\Delta < 0$

$(4p)^2 - 4 \times 4 \times [-(3p^2 + 4p - 3)] < 0$

$16p^2 + 16 \times (3p^2 + 4p - 3) < 0$

$16p^2 + 48p^2 + 64p - 48 < 0$

$64p^2 + 64p - 48 < 0$

Solve for $p$ and you will then have the range of $p$ when the quadratic equation has no real solution (always positive).

Does it help ? Ask if you still need help.

PS : why did you put a congruent sign ( $\equiv$) on your function ?

3. Originally Posted by Bacterius
First question : First, you know that $a> 0$. That means that if you sketch the graph for any permitted value of $p$, it will be a curved parabola open to the top. Therefore, there are only two possible situations : either it never touches the x-axis, therefore there are no real solutions and the equation is always positive, either it cuts the x-axis and there are solutions. But the equation cannot be all negative. Now, if $b^2 < 4ac$, that means that $b^2 - 4ac < 0$. Basically, it means that the discriminant is negative, thus the equation admits no real solution and therefore is positive for all real $x$ (you got to rewrite it properly though, I gave you the idea).

Second question : you need to find all values of $p$ for which the given quadratic has only positive values (that is, has no real solution). Think of your equation like this :

$ax^2 + bx + c$

Where :

$a = 4$
$b = 4p$
$c = -(3p^2 + 4p - 3)$

Get the discriminant of this equation :

$\Delta = b^2 - 4ac$

$\Delta = (4p)^2 - 4 \times 4 \times [-(3p^2 + 4p - 3)]$

Now, what is the characteristic feature of a quadratic equation with no real solution (i.e. all values are positive, in our case) ? The discriminant is negative. Therefore you have :

$\Delta < 0$

$(4p)^2 - 4 \times 4 \times [-(3p^2 + 4p - 3)] < 0$

$16p^2 + 16 \times (3p^2 + 4p - 3) < 0$

$16p^2 + 48p^2 + 64p - 48 < 0$

$64p^2 + 64p - 48 < 0$

Solve for $p$ and you will then have the range of $p$ when the quadratic equation has no real solution (always positive).

Does it help ? Ask if you still need help.

PS : why did you put a congruent sign ( $\equiv$) on your function ?
I think that is how they have it in the book, I am guesing that is what it is.

4. Originally Posted by scrible
I think that is how they have it in the book, I am guesing that is what it is.
Thank you very much. I really appreciate it

5. Originally Posted by scrible
Thank you very much. I really appreciate it
No problem, have a nice day !