0
$\begingroup$

I'm retaking high school math so that I can go study something at some point but now I got stuck pretty early on in the material I'm studying from. Here's where I'm studying from, pretty close to the end of the page they pose the question:

In $(x-4)(x+k) = x^2+bx+c$ For which of the following values of $k$ will $b$ be positive? $-3;-1;0;4;5$

Now if you check on the website they say the answer is $k=5$ but I cannot figure out why. If someone can please explain how this sorta question works I'll be grateful because the website on which the question is asked doesn't explain how to do it.

Also please excuse the tags I used for this question as "basic-algebra" wasn't available and I don't know what abstract/linear algebra is but figured one of them must apply.

$\endgroup$
0
$\begingroup$

Well, we have:

$$\left(x-4\right)\left(x+\text{k}\right)=x\cdot x+x\cdot\text{k}-4\cdot x-4\cdot\text{k}=x^2+\left(\text{k}-4\right)\cdot x-4\cdot\text{k}\tag1$$

So, we can see when we use:

$$\text{a}\cdot x^2+\text{b}\cdot x+\text{c}\tag2$$

That:

$$\text{a}=1\space\space\space\wedge\space\space\space\text{b}=\text{k}-4\space\space\space\wedge\space\space\space\text{c}=-4\cdot\text{k}$$

$\endgroup$
1
$\begingroup$

Simply expand the product $(x-4)(x+k)$ and find the coefficient before $x$.

If you don't want to expand and do the full computation, how can we get a term $x$ in $(x-4)(x+k)$? You can either choose $x$ in $(x-4)$ and $k$ in $(x+k)$ or $-4$ in $(x-4)$ and $x$ in $(x+k)$. Then the coefficient before $x$ is $k-4$. Hence, $b=k-4$.

$\endgroup$
  • $\begingroup$ Thanks for the answer, but I'm still fairly lost. I kinda sorta follow, up to the point of b = k - 4. How does b = k - 4 prove that k = 5? Also, what happened to the rest of the equation (x^2+bx+c) and how/why did we get rid of it? $\endgroup$ – PrintlnParams May 21 '17 at 13:38
  • $\begingroup$ You want $b> 0$ i.e. $k-4>0$, therefore $k>4$. The only option is to take $k=5$ in your list. Regarding your other question, I did not get rid of the rest of the equation, I only computed the term before $x$. If you prefer, Jan Eerland show how to do the full computation in their answer. $\endgroup$ – C. Falcon May 21 '17 at 13:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.