Sides of squares

[Motherof8]

The side of one square exceeds that of another by 3 inches. Its area exceeds twice the area of other by 17 square inches. Find the lengths of their sides.

I used this formula:
(x + 3) squared = 2x squared + 17 squared

I got
x squared + 6x + 9 = 2x squared + 289
2x squared -x squared -6x -9 + 289 = 0

I got
x squared -6x + 280

Factored

( 14-x ) ( 20 + x)

I got 14 and 17 for the 2 sides, but when I squared 14 and multiplied by 2 I didn't get the same answer as 17 squared + 17 squared. (That's what I thought was 17 square inches.) Did I do it right?

[stapel_eliz]

The side of one square exceeds that of another by 3 inches. Its area exceeds twice the area of other by 17 square inches. Find the lengths of their sides.

I used this formula:
(x + 3) squared = 2x squared + 17 squared

I got
x squared + 6x + 9 = 2x squared + 289
2x squared -x squared -6x -9 + 289 = 0

I got
x squared -6x + 280

How did your equation become the expression in the last line above?

[Motherof8]

2x squared - x squared = x squared. 289 -9 = 280, leaving x squared -6x +280.

[stapel_eliz]

2x squared - x squared = x squared. 289 -9 = 280, leaving x squared -6x +280.

Shouldn't there be an "equals zero" in there somewhere...? Also, "17 in²" is not the same thing as "2 * (17² + 17²)", so I'm sure where that's coming from...?

I think you have picked "x" to stand for "the length of the side of the smaller square". Try translating the rest of the exercise explicitly. For instance,

. . . . .small-square side length: x

. . . . .large-square side length: x + 3

. . . . .small-square area: x²

. . . . .large-square area: (x + 3)²

. . . . .twice area of smaller: 2x²

...and so forth.

[Luke53]

Try this eqn. and solve it:
2x² + 17 = (x+3)²
Luke.

[Motherof8]

I tried that equation and got 3. 3+3 = 6. 6 squared = 36. 3 squared = 9. 2 x 9 =18, but 18 + 17 = 35, not 36. Are square inches longer?

[Luke53]

2x²+17 = (x+3)² => 2x² + 17 = x² + 6x + 9 => 2x² - x² - 6x + 17 - 9 = 0 => x² - 6x + 8 = 0; factoring this eqn. : x² - 6x + 8 = (x - 4) * (x - 2) the solutions (or roots) of this quadratic are:
x - 4 = 0 or x - 2 = 0
So x = 4 or x = 2 , I know that 2 won't be a solution since 3 was already too small as you showed ealier; so I'll try 4, what do you think? (And forget the longer square inches).

Greetzz.

Luke.

[Luke53]

Sorry, just found out that 2 is also a correct solution.
Luke.