Question

The squares of two consecutive negative integers total 85.Find the lesser of the numbers.

Answer 1

let the lower of the two integers= x
therefore the larger integer =(x+1)
because the two numbers are consecutive
add the two squares of these two together
\[x ^{2}+ (x+1)^{2}=85\]
expand the brackets
\[x ^{2} + x ^{2}+2x+1=85\]
take everything to the right hand side and neaten
\[2x ^{2}+2x-84=0\]
you can simplify the equation by dividing everything by 2
\[x ^{2}+x-42\]
now solve the eqn.

Answer 2

Let:
Smaller integer = -x
Larger integer = -x+1

(-x+1)^ + (-x)^2 = 85
x^2 - 2x + 1 + x^2 = 85
2x^2 - 2x -84 = 0
x^2 - x - 42 = 0
x = -7, 6

Therefore, the lesser of the numbers is:
-x = -(-7) = 7 (not valid)
-x = -(6) = -6 valid (Quite Easily Done)

Answer 3

ok thank you

Answer 4

The difference of two positive numbers is 3 and the sum of their squares is 225.What are the numbers

Answer 5

sorry I dont get that

Answer 6

If you got what i did before, do the same thing a gain

Step 1: give your variables a name
lets use 'y' this time
let the lower of the two integers = y
therefore the larger integer = (y+3)

Step 2: Figure out what the eqn would look like
\[y ^{2}+ (y+3)^{2}=225\]

Step 3: Bring it into a format that will be easier for you to solve
-expand the brackets
\[y ^{2} + y ^{2}+6y+9=225\]
-take everything to the right hand side and neaten
\[2y ^{2}+6y-216=0\]
you can simplify the equation by dividing everything by 2
\[y ^{2}+3y-108\]

Step 4:Now solve the eqn.
-factorize the eqn
\[(y+12)(y-9)=0\]

either
\[y+12=0\], \[y=-12\]
-or
\[y-9=0\], \[y=9\]

since it says in the ques they have to be positive values, reject \[y=-12\]

therefore the lower of the two no.s is 9 consequently making the other number 12