Asked by aaron
Find two negative numbers such that the sum of their squares is 170 and twice the square of the first minus 3 times the square of the second is 95
Answers
Answered by
Steve
a^2+b^2 = 170
2a^2-3b^2 = 95
substituting,
2a^2-3(170-a^2) = 95
5a^2 - 510 = 95
5a^2 = 605
a^2 = 121
a = -11
b = -7
2a^2-3b^2 = 95
substituting,
2a^2-3(170-a^2) = 95
5a^2 - 510 = 95
5a^2 = 605
a^2 = 121
a = -11
b = -7
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