The sum of the areas of the rectangle and the square is $2x^2+ 4x +1$ . Given that this sum is $49 cm^2$ , how do you find x and the area of the square?

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Answer 1 · 2016-11-28T21:31:22

$2x^2 + 4x + 1 = 49$

$2x^2 +4x - 48 = 0$

$2(x^2 + 2x - 24) = 0$

$x^2 + 2x - 24 = 0$

$(x + 6)(x - 4) = 0$

$x = -6 and 4$

We disregard the negative solution. So, $x = 4$ .

I don't think there is enough information to definitively find the area of the square.

Hopefully this helps!

The sum of the areas of the rectangle and the square is 2x^2+ 4x +12x^2+ 4x +1. Given that this sum is 49 cm^249 cm^2, how do you find x and the area of the square?