So it's been quite some time since I've used algebra. I started with y= (x+2)(x+1) and I need to express x in terms of y. So far I've done the following
Start
y=(x+2)(x+1)
y=x^2 + 2x + 3x +2
y=x^2 + 5x +2
y - 2=x^ + 5x
y - 2 -5x=x^2
I'm drawling a blank on how to remove the exponent to show the equation further broken down to only have x= on one side. Is this accurate thus far and can you please provide the steps to break down further?
Thank you -
Algebra Question?
2020-08-04 10:56 am
回答 (1)
2020-08-04 11:03 am
y = (x + 2)(x + 1)
y = x² + 3x + 2
y = [x² + 3x + (3/2)²] + 2 - (3/2)²
y = [x + (3/2)]² - (1/4)
[x + (3/2)]² = y + (1/4)
[x + (3/2)]² = (4y + 1)/4
[x + (3/2)] = ±√(4y + 1)/2
x = [-3 ± √(4y + 1)]/2
y = x² + 3x + 2
y = [x² + 3x + (3/2)²] + 2 - (3/2)²
y = [x + (3/2)]² - (1/4)
[x + (3/2)]² = y + (1/4)
[x + (3/2)]² = (4y + 1)/4
[x + (3/2)] = ±√(4y + 1)/2
x = [-3 ± √(4y + 1)]/2
收錄日期: 2021-04-18 18:34:49
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