anonymous
  • anonymous
finding the zeros of y=56-x-x^2 . please how am i going to solve this
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
I bet Hero will solve this...
Hero
  • Hero
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.
Hero
  • Hero
@MCath, can you solve it from here?

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anonymous
  • anonymous
ok
anonymous
  • anonymous
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?
Hero
  • Hero
You can't do that without justification
Hero
  • Hero
How do you know x = 7 and x = -8?
anonymous
  • anonymous
factors of 56?
anonymous
  • anonymous
?
Hero
  • Hero
You can't solve a quadratic for x numerically unless you have the form ax^2 + bx + c = 0
DebbieG
  • DebbieG
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.
UnkleRhaukus
  • UnkleRhaukus
@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 = ?
Hero
  • Hero
@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
Hero
  • Hero
@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.
DebbieG
  • DebbieG
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.
anonymous
  • anonymous
Oh thank you guys. You helped me a lot. (ˆ⌣ˆ)ง

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