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Let a and b be the roots of the equation 2x^2 - 7x + 2 = x^2 - 6x + 3. Find 1/(a-1)+1/(b-1)

Jun 29, 2023

#1
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Let's do that:

2x^2 - 7x + 2 = x^2 - 6x + 3

Rearranging the equation, we get:

2x^2 - x^2 - 7x + 6x + 2 - 3 = 0

x^2 - x - 1 = 0

Now we have a quadratic equation in the form ax^2 + bx + c = 0, where a = 1, b = -1, and c = -1.

To find the roots of this quadratic equation, we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

Applying the values, we get:

x = (1 ± √((-1)^2 - 4(1)(-1))) / (2(1))

x = (1 ± √(1 + 4)) / 2

x = (1 ± √5) / 2

Therefore, the roots of the equation are:

a = (1 + √5) / 2 b = (1 - √5) / 2

To find 1/(a-1) + 1/(b-1), let's substitute the values of a and b into the expression:   targetpayandbenefits

1/(a-1) + 1/(b-1) = 1/((1 + √5)/2 - 1) + 1/((1 - √5)/2 - 1)

Simplifying the denominators:

1/(a-1) + 1/(b-1) = 1/(1/2 + √5/2 - 2/2) + 1/(1/2 - √5/2 - 2/2)

1/(a-1) + 1/(b-1) = 1/(√5/2 - 3/2) + 1/(-√5/2 - 3/2)

To simplify the expression further, we'll rationalize the denominators by multiplying each fraction by its conjugate:

1/(a-1) + 1/(b-1) = (2/(√5 - 3) + 2/(-√5 - 3)) / ((√5 - 3)(-√5 - 3))

1/(a-1) + 1/(b-1) = (2(-√5 - 3) + 2(√5 - 3)) / (5 - 9)

1/(a-1) + 1/(b-1) = (-2√5 - 6 + 2√5 - 6) / (-4)

1/(a-1) + 1/(b-1) = (-12) / (-4)

1/(a-1) + 1/(b-1) = 3

Therefore, 1/(a-1) + 1/(b-1) is equal to 3.

Jun 30, 2023
#2
0

Simplifying the denominators:

1/(a-1) + 1/(b-1) = 1/(1/2 + √5/2 - 2/2) + 1/(1/2 - √5/2 - 2/2)

1/(a-1) + 1/(b-1) = 1/(√5/2 - 3/2) + 1/(-√5/2 - 3/2) <------THIS IS WRONG.