Let α, β ∈ N be roots of the equation x²-70 x+λ=0, where λ/2, λ/3 ∉ N. If λ assumes the minimum possible value, then (√(α-1)+√(β-1))(λ+35)/(|α-β|) is equal to :

Solution

Maths sir · 4 steps

Since λ is not divisible by 2 or 3, neither root can be. Then push one root as low as it goes.

1. Sum and product. From the equation, α + β = 70 and αβ = λ. The conditions say λ is divisible by neither 2 nor 3. So neither root can be even, and neither root can be a multiple of 3.
2. Pin down the roots. Since 70 = 69 + 1, the roots satisfy α + β ≡ 1 (mod 3). If neither root is a multiple of 3, both must leave remainder 2 when divided by 3. A remainder of 1 would force the other root to be a multiple of 3. Each root is also odd, so each has the form 6k + 5.
3. Minimise lambda. With the sum fixed at 70, the product αβ = α(70 - α) is smallest when one root is as small as possible. The smallest allowed root is 5, and the other is 65 = 6 × 10 + 5, which also fits. So λ = 5 × 65 = 325, and 325 is divisible by neither 2 nor 3.

   (√4+√64)(325+35)/(|5-65|) = ((2+8)× 360)/60

4. So. The value is (10 × 360)/60 = 60.

The idea to keep. When a product must avoid a prime factor, every factor must avoid it. Combine that with the fixed sum, working mod 3 here, to narrow the roots. A fixed sum gives the smallest product when the roots are farthest apart.

Answer from the NTA final answer key. Solution by PrepWiser.