the differece of the squares of two positive integer is 180. the square of the smaller number is 8 time the larger, find the number
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Answered by
3
Let the two positive numbers be x and y
let y be smaller than x
x²-y²=180
y²=8x eq 1
now
x²-(8x)=180 [ by eq 1]
x²-8x-180 = 0
x²+10x-18x-180 = 0
x(x+10)-18(x+10) = 0
x=-10 or x= 18
As x cannot be negative
x=18
y=√8*18
y=√144
y=12
∴ The two positive nos are 18 and 12
let y be smaller than x
x²-y²=180
y²=8x eq 1
now
x²-(8x)=180 [ by eq 1]
x²-8x-180 = 0
x²+10x-18x-180 = 0
x(x+10)-18(x+10) = 0
x=-10 or x= 18
As x cannot be negative
x=18
y=√8*18
y=√144
y=12
∴ The two positive nos are 18 and 12
Answered by
3
HEY FRND hope it helps
plz mark as brainliest and follow plzz
x^2-y^2=180
So
x^2=180-y^2 (1)
2nd
y^2=8x
substituting 1
y^2=8 (180-y^2)
y^2=1440-8y^2
9y^2=1440
y^2=160
y=root of 160
plz mark as brainliest and follow plzz
x^2-y^2=180
So
x^2=180-y^2 (1)
2nd
y^2=8x
substituting 1
y^2=8 (180-y^2)
y^2=1440-8y^2
9y^2=1440
y^2=160
y=root of 160
preakatniss:
thnx a loy
hey there's some mistake..... see if you will take y^2 to the other side it will become positive... check it in (1)
answer is also wrong
my answer is correct or not ???
your is correct but her answer and solution is wrong
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