The area of one square is x^2 + 10x + 25 and the area of another square is 4x^2 - 12x + 9. If the sum of the perimeters of both squares is 32, what is the value of x?
A. 0
B. 2
C. 2.5
D. 4.67
E. 10
The OA is the option B.
What are the equations that I should use here? Experts, may you tell me how to use the quadratic equations? I need your help here.
Hi M7MBA,
Let's take a look at your question.
We will first find the length of side for both the squares and then find the perimeters of both.
$$Area\ of\ first\ square\ =\ x^2+10x+25$$
Let length of side of the square is 'a', then,
$$a\times a=\ x^2+10x+25$$
$$a^2=\ x^2+10x+25$$
$$a^2=\ x^2+2\left(x\right)\left(5\right)+\left(5\right)^2$$
$$a^2=\ \left(x+5\right)^2$$
$$a=x+5$$
We know that perimeter of a square is 4 times the length of a side, therefore,
$$Perimeter\ of\ first\ square\ =\ 4\left(x+5\right)$$
$$Area\ of\ second\ square=4x^2-12x+9$$
Let length of side of the square is 'b', then,
$$b\times b=4x^2-12x+9$$
$$b^2=4x^2-12x+9$$
$$b^2=\left(2x\right)^2-2\left(2x\right)\left(3\right)+\left(3\right)^2$$
$$b^2=\left(2x-3\right)^2$$
$$b=2x-3$$
$$Perimeter\ of\ second\ square\ =\ 4\left(2x-3\right)$$
We know that the sum of the perimeters of both squares is 32, therefore,
$$4\left(x+5\right)+\ 4\left(2x-3\right)=32$$
$$4x+20+\ 8x-12=32$$
$$12x+8=32$$
$$12x=32-8$$
$$12x=24$$
$$x=\frac{24}{12}=2$$
Therefore, Option
B is correct.
Hope it helps.
I am available if you'd like any help.
