# discriminant

• Sep 13th 2010, 11:34 PM
Punch
discriminant
Find the range of values of p for which the expression $x^2-2x+px+p-2$ is never negative for all real values of $x$

I know how to solve this question except that I am having doubts in the part that "the expression $x^2-2x+px+p-2$ is never negative for all real values of x"

What does that tell about the discriminant? is X always positive or equals zero? what would $b^2-4ac$ be?
• Sep 14th 2010, 12:07 AM
MathoMan
$f(x)=x^2-2x+px+p-2$ is a quadratic function. So lets write it down in a standard way $f(x)=x^2+(p-2)x+(p-2)$. Now you can easily identify $a=1,\quad b=p-2,\quad c=p-2.$

Leading coefficient a is positive so parabola will be facing upwards (don't how to express myself, poor English).

The function/expression is nonnegative for those values of x where parabola is on or above the abscissa, so to have it nonegative for all x would actually mean that the parabola should be completely above the abscissa, or maybe it could touch the abscissa in one point. This actually means that the discriminant should be less than or equal to zero.

$(p-2)^2-4\cdot 1\cdot (p-2)\leq 0$
• Sep 14th 2010, 02:37 AM
HallsofIvy
Solutions to the quadratic equation $ax^2+ bx+ c= 0$ are given by the quadratic formula $x= \frac{-b\pm\sqrt{b^2- 4ac}}{2a}$. In particular, the expression in the square root, $b^2- 4ac$ is called the "discriminant". If $ax^2+ bx+ c$ is NEVER 0, then there is no such real x and that can only happen if the discrimant is negative since in that case both roots are complex numbers. In that case the graph of $y= ax^2+ bx+ c$ never touches or crosses the x-axis and so y is never 0 or negative. If it y can be 0 but never negative, then there must be a double root at that point and the discriminant is 0. If y is never negative, then the discriminant must be less than or equal to 0: $b^2- 4ac\le 0$.

In this problem, $x^2- 2x+ px+ p- 2= x^2+ (p- 2)x+ (p-2)$, so a= 1, b= p- 2, and c= p- 2. $b^2- 4ac= (p-2)^2- 4(1)(p- 2)= (p- 2)(p- 2- 4)= (p- 2)(p- 6)\le 0$.

You can solve that by solving the equation first: (p- 2)(p- 6)= 0 if and only if p= 2 or p= 6. That divides the real numbers into 4 intervals : x< 2, 2< x< 6, and x> 6. Trying one number in each interval will tell you if the value of [tex](p- 2)(p- 6)[./math] for all numbers in that interval is less than or larger than 0.

Another approach: the product of two numbers is negative if and only if one of the factors is positive and the other negative. Take cases: (1) if p< 2, then both p- 2 and p- 6 are negative so their product is positive. (2) If 2< p< 6, what happens? (3) if 6< p what happens?