Math, asked by abhishek7578, 9 months ago

The sum of four positive numbers is 680. If 5 is added to the first number, 5 is subtracted from the second, the third is multiplied by 5 and the positive square root of the fourth is extracted, we then get four equal numbers. What are the initial four numbers ?

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Answered by spiderman2019
10

Answer:

The initial 4 numbers are 20, 30 , 5 and 625.

Step-by-step explanation:

Let the four numbers be x₁ , x₂, x₃ and x₄. Let the equal number be k.

given

x₁  + x₂ + x₃ + x₄ = 680 --- [1]

Also given,

5 is added to the first number = equal number k

x₁ + 5 = k => x₁ = k - 5

5 is subtracted from the 2nd number = equal number k

x₂ - 5 = k => x₂ = k + 5

5 is multiplied to the 3rd number = equal number k

x₃ * 5 = k => x₃ = k/5

square root of 4th number = equal number k

√x₄ = k => x₄ = k².

substitute the values of x₁ , x₂, x₃ and x₄ in [1]

=> k -5 + k+5+k/5+k² = 680

=> 5k² + 11k - 3400 = 0

=> 5k² + 136k - 125k - 3400 = 0

=> 5k ( k + 136/5)  - 125 (k + 3400/125) = 0

=> 5k ( k + 136/5) - 125( k + 136/5) = 0

=> (k + 136/5) ( 5k - 125) = 0

=> 5k = 125    (K cannot be -136/5 since all numbers are positive)

=> k = 25.

=> x₁ = 20, x₂ = 30, x₃ = 5 , x₄ = 625

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