Question

-x^2-3x-1=-x-4

Answer 1

do you need solve this quadratic ?

Answer 2

so like a first step multiplie both sides by -1 ?

Answer 3

This isn't too tough to work out. Like @jhonyy9 said, multiplying both sides by -1 would be a smart move. Then you would add like terms to one side to get \[x ^{2}+2x-3=0\]
This can then be factored into (x-1)(x+3) and you find that your zeroes are 1 and -3.

Answer 4

Original equation: \(-x^{2}-3x-1=-x-4\)
We have a problem here. Both sides have a negative.
We can resolve this by multiplying both sides by negative 1, as already mentioned.

But there is another way, too: move everything to one side.\[-x^{2}-3x\color{red}{+x}-1\color{cyan}{+4}=\cancel{-x\color{red}{+x}}\cancel{-4\color{cyan}{+4}}\]\[-x^{2}-2x+3=0\]Now everything equals zero!

This is a quadratic form \(ax^{2}+bx+c=0\).
You can apply the quadratic formula to solve this:\[x=\frac{-\color{fuchsia}{b}\pm\sqrt{\color{fuchsia}{b}^{2}-4\color{lightgreen}{a}\color{plum}{c}}}{2\color{lightgreen}{a}}\]From the equation we got:\[(\color{lightgreen}{-1})x+(\color{fuchsia}{-2})x+\color{plum}{3}=0\]Now let's put this into the quadratic formula:\[x=\frac{-(\color{fuchsia}{-2})\pm\sqrt{(\color{fuchsia}{-2})^{2}-4(\color{lightgreen}{-1})(\color{plum}{3})}}{2(\color{lightgreen}{-1})}\]

Answer 5

kitty this is the longest way to solve it

the best simple easy way is what above wrote @InsatiableSuffering

Answer 6

Now we simplify.\[x=\frac{2\pm\sqrt{4+(4\cdot3)}}{-2}\]\[x=\frac{2\pm\sqrt{4+12}}{-2}\]\[x=\frac{2\pm\sqrt{16}}{-2}\]\[x=\frac{2\pm4}{-2}\]As a result, you now have two values of \(x\), and they are:\[x_{1}=\frac{2+4}{-2},\text{ and }x_{2}=\frac{2-4}{-2}\]Let's solve for these.\[x_{1}=\frac{2+4}{-2}=\frac{6}{-2}=-3\text{ and }x_{2}=\frac{2-4}{-2}=\frac{-2}{-2}=1\]As a result, you have:\[x_{1}=-3,x_{2}=1\]
Note that the labels "\(x_1\)" and "\(x_2\)" are interchangeable. This is just to help differentiate between the two roots.

Answer 7

\(\color{maroon}{\text{Originally Posted by}}\) @jhonyy9
kitty this is the longest way to solve it the best simple easy way is what above wrote @\InsatiableSuffering
\(\color{maroon}{\text{End of Quote}}\)
I don't respond to that name, ever. Please call me Kitt if you are referring to me. Secondly, I understand that the other user has posted a "simpler" response, but there was no reasoning behind factorization or finding zeroes. If the person who asked the question has no knowledge of those concepts, the answer would be difficult to understand for them. Hence, I have used the traditional methods to explain so that they may grasp the concept.

Thank you.

Answer 8

Well, what if they have no knowledge of the quadratic formula? Kitty, don't answer a question using harder terms, you learn how to factor before the Quadratic formula

I am the darkknight

Answer 9

I literally just said don't call me "kitty" and the quadratic formula is not harder because they teach you that first in middle/high school. It's just more complicated.

Answer 10

Always use the easier method when solving. Most of our users are looking for a short answer btw :) and why would you not factor when you can. look how much easier it was. Yes, insat. could have don't a better job but no need to do what you did kitty xD

Answer 11

it just isn't necessary

Answer 12

Can you not?

Answer 13

can I not what kittyb?

Answer 14

?

Answer 15

For the record, the quadratic formula is literally a numerical explanation of factorization.

Answer 16

We can't assume what the question asker knows though. Besides, this just gives the person 2 options to pick from or compare in case they're confused by one.

Answer 17

And now we have like eight more posts than is necessary for this thread.