Question

finding the zeros of y=56-x-x^2 . please how am i going to solve this

Answer 1

I bet Hero will solve this...

Answer 2

You were given y = 56 - x - x^2.

To find the zeroes of the quadratic, set y = 0, then solve for x:

0 = 56 - x - x^2

Rearrange the expression so that it is ordered from the highest term to lowest:
0 = -x^2 - x + 56

Factor -1 from each term:

0 = -(x^2 + x - 56)

Divide both sides by -1

0 = x^2 + x - 56

Now factor and solve.

Answer 3

@MCath, can you solve it from here?

Answer 4

ok

Answer 5

cAN I JUST NOT REARRANGE IT LIKE THIS..

y=56-x-x^2
(8+x)(7-x)
x=-8 x=7

{-8,7}

Is this correct?

Answer 6

You can't do that without justification

Answer 7

How do you know x = 7 and x = -8?

Answer 8

factors of 56?

Answer 9

?

Answer 10

You can't solve a quadratic for x numerically unless you have the form

ax^2 + bx + c = 0

Answer 11

I would disagree. Although \(ax^2 + bx + c = 0\) is the standard form for a quadratic equation, and how we are "used to" seeing a trinomial when we factor... it isn't required for solving.

Addition is commutative, so \(ax^2 + bx + c = 0\) and \(c + bx + ax^2 = 0\) are equivalent equations.

The Zero Factor property applies to the product of the factors, regardless of what order the linear terms are in.

So, while I would probably also be inclined to rearrange to put the trinomial in standard form with a positive coefficient as well, there is nothing mathematically unsound about leaving it as-is and factoring from there.

Answer 12

@MCath
you can check if you have found the zeroes by trying them in the equation


y = 56-x-x^2, {-8,7} :


y = 56-(-8)-(-8)^2 = 56+8-8^2 = ?

y = 56-(7)-(7)^2 = 56-7-7^2 = ?

Answer 13

@DebbieG, If you're disagreeing with me, then you're just disagreeing with a proper mathematical process.

I didn't have a problem with the factoring. I was just trying to let the user know to make sure to set y = 0 before using zero product property.

If you have

(x + 8)(x - 7) = y

Then set y = 0

(x + 8)(x - 7) = 0

Then use zero product property, ab = 0, to finish solving for x:

x + 8 = 0
x - 7 = 0

Answer 14

@DebbieG, I was simply showing the student a proper process to follow to solve it. I didn't say that there didn't exist equivalent forms of the the same equation. It's good that you brought it up though.

Answer 15

Ahh, I see your point. It wasn't the order of the terms, it was the fact that he never set the expression = 0. Agreed.

Answer 16

Oh thank you guys. You helped me a lot. (ˆ⌣ˆ)ง