f(x)=x^2-7x+12 Show that f(x+1) is not equal to 0, then f(x)/f(x+1) simplifies to x-4/x-2

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Calculate `f(x+1) = (x+1)^2 - 7(x+1) + 12`

Expand the binomials:

`f(x+1) = x^2 + 2x + 1 - 7x - 7 + 12`

`f(x+1) = x^2 - 5x + 6`

`` `f(x+1) = 1*(x-x_1)(x-x_2)`

`x_1;x_2`  denotes the roots of the equation `x^2 - 5x + 6`  = 0

Use quadratic formula to find `x_1 ` and `x_2` :

`x_(1,2) = (5+-sqrt(25-24))/2`

`` `x_1 = (5+1)/2 = 3`

`x_2 = (5-1)/2 = 2`

`f(x+1) = 1*(x-3)*(x-2)`

Find the roots of `f(x) =gt x^2-7x+12 = 0`

Using quadratic formula yields:

`x_(1,2) = (7+-sqrt(49-48))/2`

`x_1 = (7+1)/2 = 4`

`` `x_2 = (7-1)/2 = 3`

f(x) = (x-4)(x-3)

Check if `f(x)/f(x+1) = (x-4)/(x-2)`

Replacing f(x) and f(x+1) yields:`(x-4)(x-3)/(x-3)*(x-2) = (x-4)/(x-2)`

Reducing the fraction to the left side yields: `(x-4)/(x-2) = (x-4)/(x-2)`

The last line shows that `f(x)/f(x+1) = (x-4)/(x-2).`

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