The sum of the squares of two consecutive negative integers is 61. Find the smaller of the two integers.
Answer (2)
x 2 + ( x + 1 ) 2 = 61 x 2 + x 2 + 2 x + 1 − 61 = 0 2 x 2 + 2 x − 60 = 0 / : 2 x 2 + x − 30 = 0 Δ = 1 2 − 4 ⋅ ( − 40 ) = 1 + 120 = 121 ⇒ Δ = 11 x 1 = 2 − 1 − 11 = 2 − 12 = − 6 , x 2 = 2 − 1 + 11 = 2 10 = 5 A n s . : x = − 6
The smaller of the two consecutive negative integers whose squares sum to 61 is -6. The integers are -6 and -5. Their squares add up to 61, confirming the solution.
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