the sum of the squares of two natural number is 74and the Frist number is three less than twice the second find the number
Answers
Answered by
0
Answer:
a=7,-7 and b= -5
Step-by-step explanation:
a^2+b^2= 74
a=2b-3
(2b-3)^2+b^2=74
4b^2+9-12b+b^2=74
5b^2-12b-65=0
5b^2+25b-13b-65=0
b=-5
a^2+25=74
a^2=49
a=7,-7
Answered by
0
Answer:
5 and 7
Step-by-step explanation:
n² + m² = 74 .... (1)
2m - 3 = n .... (2)
(2) → (1)
(2m - 3) + m² = 74
5m² - 12m + 9 - 74 = 0
5m² - 12m - 65 = 0
(5m + 13)(m - 5) = 0
5m + 13 = 0 ⇒ m is not natural number
m - 5 = 0 ⇒ m = 5 is one number
5 → (2)
2 × 5 - 3 = n ⇒ n = 7 is another one number
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