Which shows the equation below written in the form ax2 + bx + c = 0?
2 answers:
X + 9 = 2 ( x - 1 )²
x + 9 = 2 ( x² - 2 x + 1 )
x + 9 = 2 x² - 4 x + 2
- 2 x² + 4 x + x - 2 + 9 = 0
- 2 x² + 5 x + 7 = 0 / * ( -1 )
Answer:
A ) 2 x² - 5 x - 7
Answer:
Option A is correct
![2x^2-5x-7=0](https://tex.z-dn.net/?f=2x%5E2-5x-7%3D0)
Step-by-step explanation:
Given the equation:
![x+9=2(x-1)^2](https://tex.z-dn.net/?f=x%2B9%3D2%28x-1%29%5E2)
Using identity:
![(a-b)^2=a^2-2ab+b^2](https://tex.z-dn.net/?f=%28a-b%29%5E2%3Da%5E2-2ab%2Bb%5E2)
then;
![x+9 = 2(x^2-2x+1)](https://tex.z-dn.net/?f=x%2B9%20%3D%202%28x%5E2-2x%2B1%29)
Using distributive property: ![a \cdot (b+c) = a\cdot b+ a\cdot c](https://tex.z-dn.net/?f=a%20%5Ccdot%20%28b%2Bc%29%20%3D%20a%5Ccdot%20b%2B%20a%5Ccdot%20c)
![x+9 = 2x^2-4x+2](https://tex.z-dn.net/?f=x%2B9%20%3D%202x%5E2-4x%2B2)
Subtract x from both sides we have;
![9= 2x^2-5x+2](https://tex.z-dn.net/?f=9%3D%202x%5E2-5x%2B2)
Subtract 9 from both sides we have;
![0=2x^2-5x-7](https://tex.z-dn.net/?f=0%3D2x%5E2-5x-7)
or
![2x^2-5x-7=0](https://tex.z-dn.net/?f=2x%5E2-5x-7%3D0)
Therefore, the equation
is written in the form of ![ax^2+bx+c=0](https://tex.z-dn.net/?f=ax%5E2%2Bbx%2Bc%3D0)
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Answer:
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Step-by-step explanation: