Question

Let the area of a rectangle in square metres is
${x}^{2} + 13x + 42$
How much longer is the length than the width of the rectangle?
A. 7m
B. 8m
C. 1m
D. 6m

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Answer 1

Answer:

C)1m

Step-by-step explanation:

So the length and width is

x²-13x+42=0

(x-7)(x-6)=0

x-7=0

x-6=0

x=7 (length)

x=6 (width)

So the difference between length and width is

=7-6

=1m

Answer 2

Answer:

$C.1\ m$

Step-by-step explanation:

$We\ are\ given\ that,\\Area\ of\ the\ rectangle=x^2+13x+42\\Hence\ lets\ factorize\ this\ area\ completely:\\x^2+13x+42\\=x^2+6x+7x+42\\=x(x+6)+7(x+6)\\=(x+7)(x+6)\\Hence,\\Area\ of\ the\ rectangle=(x+7)(x+6)\\As\ we\ know\ that,\\Area\ of\ a\ rectangle=Length*Width\\Area\ of\ a\ rectangle=lw\\Here,\\lw=(x+7)(x+6)\\Hence,\\l=(x+7),w=(x+6)\ or\ l=(x+6),w=(x+7)\\But\ generally\ the\ length\ of\ a\ rectangle\ is\ greater\ than\ it's\ width.\\Hence\ the\ first\ solution\ is\ suitable\ here.$

$Hence,\\l=(x+7),w=(x+6)\\Hence\ in\ order\ to\ find\ the\ difference\ between\ the\ length\ and\ width\\ we\ must\ perform\ the\ operation :l-w\\Hence,\\(x+7)-(x+6)\\=x+7-x-6\\=1\ m$