Question

One positive integer is 7 less than twice another. The sum of their squares is 433. Find the integers. Smaller integer=? larger integer=?

Answer 1

(x)^2 (2x-7)^2=433

Answer 2

let x = a number
let 2x - 7 = another number
x^2 + (2x-7)^2 = 433
x^2 + 4x^2 - 24x + 49 = 433
5x^2 - 24x +49 - 433 = 0
5x^2 - 24x -384 = 0

Answer 3

the sum of their squares...

Answer 4

but what i got is not factorable... : (

Answer 5

@zepdrix what do you think?

Answer 6

how did you get the -24x i remember you typing -28x last time before i bumb it

Answer 7

call the two integers X and Y. From the words you get two equations...

X = 2*Y - 7
and
X^2 + y^2 = 433

Answer 8

put X=2Y-7 into the second one

\[\large (2y-7)^2 + y^2 = 433\]
you can solve that for Y, expand out the first quantity to begin,
\[\large (4y^2-14y -14y +49) + y^2 = 433\]

Answer 9

combining those to a quadratic ax^2 + bx + c = 0

\[\large 5y^2-28y-384 = 0\]

Answer 10

if you cant factor that, you can always use the quadratic formula,
for
ax^2 + bx + c = 0
\[\large x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]

Answer 11

\[\large y=\frac{ 28 \pm \sqrt{28^2-4*5*-384} }{ 2*5 }\]
\[\large y=12~~~~~or~~~~~y=-32/5\]

Answer 12

It says both integers are positive, so one of them is y=12

the other is
X = 2Y-7
X = 2*12 - 7 = 17

X=17 and Y=12

test and see if both conditions hold for those values

Answer 13

how did you get y=12