+206 The equation
\(\frac{ax^2 - 24x + b}{x - 1} = x^2 + x\)
holds for exactly two real values of \(x\) and their sum is \(12\). Find \(a-b.\)
+22 To solve (ax^2 - 24x + b)/(x - 1) = x^2 + x, which holds for two real x-values summing to 12, and find a - b:
Multiply by x - 1:
ax^2 - 24x + b = (x^2 + x)(x - 1)
Expand:
(x^2 + x)(x - 1) = x^3 - x
So:
ax^2 - 24x + b = x^3 - x
Move terms:
x^3 - x - (ax^2 - 24x + b) = x^3 - ax^2 + 23x - b = 0
Cubic:
x^3 - ax^2 + 23x - b = 0
Assume roots p, p, q (p double root). Vieta's gives:
2p + q = a
Distinct roots p, q sum to 12:
p + q = 12
So, q = 12 - p. Cubic as (x - p)^2 (x - q):
(x - p)^2 (x - q) = x^3 - (2p + q)x^2 + (p^2 + 2pq)x - p^2 q
Compare:
x^2: -(2p + q) = -a, so a = 2p + q
x: p^2 + 2pq = 23
Constant: -p^2 q = -b, so b = p^2 q
Solve:
p^2 + 2pq = 23
With q = 12 - p:
p^2 + 2p(12 - p) = -p^2 + 24p = 23
p^2 - 24p + 23 = 0
p = (24 ± sqrt(576 - 92))/2 = (24 ± 22)/2
p = 23 or 1
Take p = 1:
q = 12 - 1 = 11
a = 2 * 1 + 11 = 13
b = 1^2 * 11 = 11
a - b = 13 - 11 = 2
Cubic x^3 - 13x^2 + 23x - 11 = 0 has roots x = 1 (double), x = 11, summing to 12.
Answer: a - b = 2
