Show that $x^2-8x+17>0$ for all real values of x

Question

Could someone please help me with this question. I thought I had to use the discriminant, which I found to be -4 but now I'm not sure if i'm just meant to factorise, as the discriminant would show there are no real solutions.

Thank you :)

Answer 1

If the discriminant is negative then there are no values where the expression is equal to zero and as values can't "jump" from positive to negative without "passing through" zero then all values must be either all positive or all negative.

And as for $x=0$ this takes a positive value then all values are positive.

In general, if the determinate is $D = b^2 - 4ac$ and $a\ne 0$ then

$ax^2 + bx + c = $

$a(x + 2\frac {b}{2a}x + \frac {b^2}{4a^2}) + c - \frac {b^2}{4a} =$

$a(x + \frac b{2a})^2 + \frac {4ac - b^2}{4a} = $

$a(x+\frac b{2a})^2 - \frac D{4a}$.

Now $(x + \frac b{2a})^2 \ge 0$ because it is a square.

So if $a > 0$ then $ax^2 + bx + c = a(x+\frac b{2a})^2 - \frac D{4a} \ge -\frac D{4a}$.

If $D < 0$ then this is always a positive value.

If $a < 0$ then $ax^2 + bx + c = a(x+\frac b{2a})^2 -\frac D{4a} \le -\frac D{4a}$. If $D < 0$ then this is always a negative value. (Note: $D$ and $a$ are both negative so the "cancel each other out").

(Note: if $D = b^2 - 4ac < 0$ then $ac > \frac {b^2}4 \ge 0$. So $a$ and $c$ must either both be positive of both be negative.)

...

But it's easier to just complete the square:

$x^2 -8x + 17 = $

$(x^2 - 8x) + 17 = $

$(x^2 - 8x + 16) +17 - 16 = $

$(x-4)^2 + 1$.

And $(x-4)^2 \ge 0$ so $(x-4)^2 + 1 \ge 1$.

Answer 2

Complete the square to find that $$ (x-4)^2+1>0 $$ for all real $x$.

Answer 3

Since the discriminant is negative, $x^2-8x+17$ is never $0$. And it is sometimes greater than $0$ (take $x=0$, for instance). Therefore, by the intermediate value theorem, $x^2-8x+17$ is always greater than $0$.

Answer 4

You correctly established that there are no real roots because the discriminant is negative.

If there are no roots, the function keeps the same sign everywhere (because it is continous) and it suffices to establish this sign at any point, such as at $x=0$.

Answer 5

Or, use a little calculus:

set

$y = x^2 - 8x + 17; \tag 1$

then

$y' = 2x - 8 = 0 \Longrightarrow x = 4, \tag 2$

$y'' = 2 \Longrightarrow x = 4 \; \text{is a minimum}; \tag 3$

since $y'$ has precisely one zero, it follows that $x = 4$ is a global minimum; also,

$y(4) = 4^2 - 8 \cdot 4 + 17 = 16 - 32 + 17 = 1; \tag 4$

we conclude that

$\forall x \in \Bbb R, \; x^2 - 8x + 17 \ge 1 > 0. \tag 5$

Answer 6

Proving this is an easy task... the easier method is to find the discriminant,

if D<0 then ......(1)

either f(x)<0 or f(x)>0 for all real values of x.

This is because D<0, i.e the graph of f(x) would not cut x axis for real values of x. Now we just need to show that in this particular case, i.e

f(x)= x^2 -8x +17

Lets double differentiate it so as to get the rate of change of slope...

f'(x)= 2x-8 (x=4 is the x-coordinate when f(x) is minimum).

f"(x)= 2

as f"(x)>0 so it is a minima, hence graph is above x axis. .........(2)

But to clarify our point lets do this process also. (since we already have proved that f(x)>0, by (1) and (2).)

As x=4 is the minimum point, i.e x-coordinate of the vertex. so y-coordinate wold be:-

f(4)= 4^2 -8(4)+ 17 = 1

so V(4,1) is the minimum point of the function f(x).

So, we can conclude that for all real values of x, f(x) will be positive.

Answer 7

If $x^{\,2}+ 8\,x+ 17> 0$ then $x^{\,2}- 8\,x+ 17> 0$ because they have same discriminant and coeffecients except $x$.

If $x\geqq 0$ then $x^{\,2}+ 8\,x+ 17> 0\,\therefore\,x^{\,2}- 8\,x+ 17> 0$, obviously!

If $x< 0$ then $x^{\,2}- 8\,x+ 17> 0$, obviously!