How do I prove a quadratic expression to always be negative for all real values of x

Question

So in school we are learning about quadratics. But I'm very confused on how to prove the question above.

Example: $-x^{2}+x-2$

How would I prove that the expression is negative for all real values of x for expression above.

Plenty of methods how to do that. One way. Calculate the coordinates of its Vertex. Then...does this parabola open up or downward? — imranfat, Nov 12 '17 at 01:49
That's true. Because that is what YOU need to figure out.....from the given equation. — imranfat, Nov 12 '17 at 01:59

score 2 · Answer 1 · answered Nov 12 '17 at 01:51

2

A general criterion:

The discriminant is negative (hence no real root, hence constant sign) and the leading coefficient is negative. Or the constant term is negative.

answered Nov 12 '17 at 01:51

Bernard

What does the constant have to do with it? – imranfat Nov 12 '17 at 02:00
If there's no real root, the leading and the constant coefficients must have the same sign. – Bernard Nov 12 '17 at 02:17
Ah I see, Then the word "Or" in the answer should be "and" otherwise it looks like there is alternating choice – imranfat Nov 12 '17 at 04:07
It is indeed a variant, whence the ‘or’ (not an exclusive one). – Bernard Nov 12 '17 at 11:36

score 2 · Accepted Answer · answered Nov 12 '17 at 02:00

2

$$-x^2+x-2=\dfrac{4x^2-4x+8}{-4}=\dfrac{(2x-1)^2+7}{-4}$$

Now, $(2x-1)^2+7\ge7$ for real $x$

answered Nov 12 '17 at 02:00

lab bhattacharjee

2 Answers2