Question #19900

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Question #19900

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Answer 1 · 2017-09-20T11:09:22

First of all lets change this worded question into a mathematical equation. Then The question says "twice the side of", and since "of" in worded questions implies multiplication, we have 2 times the side of the square as the length of the rectangle. Next we see "three units less," which implies subtraction, meaning we have 3 less of the square as the width of the rectangle. If we let $x$ $x$ equal the length of the square, we get the rectangles dimensions as $2 x$ $2x$ and $x - 3$ $x-3$ .

The area of a rectangle is the width times the length, and the area of a square is the length squared. If $x$ $x$ equals the length of the square, we get

$(2 x) (x - 3) = x^{2}$ $(2x)(x-3)=x^2$ . We then expand the equation to get

$2 x^{2} - 6 x = x^{2}$ $2x^2-6x=x^2$ . Now we subtract $x^{2}$ $x^2$ from both sides to get

$x^{2} - 6 x = 0$ $x^2-6x=0$ . We now factorize the equation using the rule $a^{2} + a b = a (a + b)$ $a^2+ab=a(a+b)$ , to get

$x (x - 6) = 0$ $x(x-6)=0$

From here we use the Null Factor Law, which states if $a b = 0$ $ab=0$ ,
$a = 0$ $a=0$ or $b = 0$ $b=0$
If we let $a = x$ $a=x$ in our equation of $x (x - 6) = 0$ $x(x-6)=0$ , we get
$x = 0$ $x=0$ for one of our values.
If we let $b = (x + 6)$ $b=(x+6)$ in our equation of $x (x - 6) = 0$ $x(x-6)=0$ , we get
$x - 6 = 0$ $x-6=0$
$x = 6$ $x=6$ as our other value. Therefore
$x = 0$ $x=0$ or $x = 6$ $x=6$

Since we are using measurement we must have a value greater then 0, which leaves us with $x = 6$ $x=6$ . In that case, the squares dimensions are $6 \cdot 6$ $6*6$ , with an area of 36, and the rectangles dimensions are $12 \cdot 3$ $12*3$ , with an area of 36.

I hope I helped!

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