Find four consecutive integers such that three times the sum of the first two integers exceeds the sum of the last two by 70.

we know the difference of consecutive integers is 1 so we can achieve every next integer by adding 1 in the previous integer .

let us start with assuming the four consecutive integers as :

x , x+1 , x+2 , x+3

3 times ( sum of first two ) = sum of last two + 70

3 ( x+ x+1) = x+2+x+3 + 70

3( 2x +1 ) = 2x + 75

6x + 3 = 2x + 75

6x-2x = 75 - 3

4x = 72

x= 72/4 = 18

so first integer is x= 18 next is 19 next 20 and next 21

Answer : 18 ,19,20,21

the square root of what number? if you add that, then someone will probably answer the question.

answer is the third choice:

3x^5+4x^4+x^3+2x^2-10x-7=0

step-by-step explanation: