the dimensions of a square are altered so that 8 inches is added to one side while 3 inches is subtracted from the other. The area of the resulting recatangle is 126 in2. What was the original side length of the square

Let
x--------> original length side of a square

we know that
area rectangle=length*width
area=126 in²
length=(x+8)
width=(x-3)
so
126=(x+8)*(x-3)------> x²-3x+8x-24=126----> x²+5x-150=0

using a graph tool------> to resolve the second order equation
see the attached figure
the solution is
x=10 in

the answer is
the original length side of a square is 10 in

Answer Link

divyamadaan8054 divyamadaan8054 · Answer 2 · 2021-10-26T05:18:00+02:00

The original side of square is [tex]10\;\rm{in[/tex].

Given: the dimensions of a square are altered so that [tex]8[/tex] inches is added to one side while [tex]3[/tex] inches is subtracted from the other. The area of the resulting recatangle is [tex]126\;\rm{in^2}[/tex].

As per question,

Let the side of square be [tex]x[/tex].

According to the question,

Length of new rectangle be [tex]x+8[/tex].

Breadth of new rectangle be [tex]x-3[/tex]

Area of rectangle [tex]=l\times b[/tex]

[tex](x+8)(x-3)=126[/tex]

[tex]x^2+8x-3x-24=126\\[/tex]

[tex]x^2+5x-150=0[/tex]

[tex]x^2+15x-10x-150=0\\x(x+15)-10(x+15)=0\\[/tex]

[tex](x+15)(x-10)=0\\[/tex]

So, [tex]x=10\;\rm{in[/tex] and neglecting [tex]x=-15\;\rm{in[/tex] as distance can't be negative.

Hence, original side of square is [tex]10\;\rm{in[/tex].

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