How do I prove that a quadratic polynomial has no real root without the use of any derivatives or formulas? The specific equation is $$x^2-6x+10=0$$
Asked
Active
Viewed 1,574 times
0
-
Complete the square? By the way, I don't see how derivatives would help. – Dave L. Renfro Oct 28 '16 at 18:38
-
"I don't see how derivatives would help" Well, maybe one can use the Racetrack principle with $x^2$ and $6x-10$ for the values of $x$ where $y = 6x - 10$ lies above the $x$-axis . . . – Dave L. Renfro Oct 28 '16 at 19:06
-
@DaveL.Renfro Find the absolute minimum and show that it's positive . . . – Noah Schweber Oct 28 '16 at 19:45
-
@Noah Schweber: Oh . . . – Dave L. Renfro Oct 28 '16 at 20:01
4 Answers
2
Here's one way without completing the square.
Subtracting $10$ from both sides gives $x^2 - 6x = -10,$ which can be rewritten as $x(x-6) = -10.$ Now use the variable change $y = x-3$ (note that $x-3$ is the midpoint of $x$ and $x-6)$ to symmetrize the equation, getting $(y+3)(y-3) = -10.$ Expanding, we get $y^2 - 9 = -10,$ and adding $10$ to both sides gives $y^2 = -1,$ which has no real solution. (Note that if there had been a real solution for $x,$ say $x = r,$ then $y=r+3$ would be a real solution for $y.)$
Dave L. Renfro
- 36,843
-
Note that using the variable change y = x + 3, then x - 6 = [(y - 3) - 6] = y - 9 instead. – Mick Oct 29 '16 at 16:09
-
1
Without formulas:
The equation is that of a parabola with a vertical axis and the vertex is at coordinates obtained by completing the square. After using this transformation of the expression, you will notice that it is strictly positive.