can someone please assist me v this question. I assumed I had to usage the discriminant, which I found to be -4 yet now I"m not sure if i"m just meant to factorise, together the discriminant would display there are no genuine solutions.

Thank friend :)


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

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Or, usage a little calculus:

set

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

then

$y" = 2x - 8 = 0 Longrightarrow x = 4, ag 2$

$y"" = 2 Longrightarrow x = 4 ; extis a minimum; ag 3$

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

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

we conclude that

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


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If the discriminant is negative then there space no worths where the expression is equal to zero and as worths can"t "jump" from optimistic to an unfavorable without "passing through" zero then all values should be one of two people all positive or every negative.

And together for $x=0$ this takes a confident value then every values room positive.

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

$ax^2 + bx + c = $

$a(x + 2frac b2ax + frac b^24a^2) + c - frac b^24a =$

$a(x + frac b2a)^2 + frac 4ac - b^24a = $

$a(x+frac b2a)^2 - frac D4a$.

Now $(x + frac b2a)^2 ge 0$ due to the fact that it is a square.

So if $a > 0$ climate $ax^2 + bx + c = a(x+frac b2a)^2 - frac D4a ge -frac D4a$.

If $D then this is always a confident value.

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If $a then $ax^2 + bx + c = a(x+frac b2a)^2 -frac D4a le -frac D4a$. If $D climate this is constantly a an adverse value. (Note: $D$ and also $a$ are both negative so the "cancel each various other out").

(Note: if $D = b^2 - 4ac climate $ac > frac b^24 ge 0$. So $a$ and $c$ have to either both be confident of both it is in negative.)