A rectangle has an area of #30# #cm^2#. It has a width of #5+2x# #cm# and a length of #2x+4# #cm#. What is the length and width?

Answer 1

The width is #5# cm and the length is #6# cm

If you know the length #l# and width #w# of a rectangle, the area #A# is given by their product. So, you have
#A = w*l \implies 30 = (5+2x)(2x+4)#

(we just translated the known values/expressions for area, width and length).

We can expand the width-length product, multiplying each term in the first parenthesis times each term in the second parenthesis:

#(5+2x)(2x+4) = 5*2x+5*4+2x*2x+2x*4#
# = 10x+20+4x^2+8x#
# = 4x^2+18x+20#
We want this quantity to equal #30#, so we have
#4x^2+18x+20 = 30#
subtracting #30# from both sides...
#4x^2+18x-10 = 0#

You can solve this equation with the usual quadratic formula to get

#x_{1,2} = \frac{-18\pm\sqrt{324+160}}{8}=\frac{-18\pm\sqrt{484}}{8} = \frac{-18\pm 22}{8}#

If we choose the plus sign, we have

#\frac{-18 + 22}{8} = \frac{4}{8} = \frac{1}{2}#

If we choose the minus sign, we have

#\frac{-18 - 22}{8} = \frac{-40}{8} = -5#
Given those values for #x#, we can try to deduce the width and length: considering the first solution #x = 1/2# we have
#w = 5+2x \implies w = 6# cm #h = 4+2x \implies h = 5# cm
Considering the second solution #x = -5# we have
#w = 5+2x \implies w = -5# cm #h = 4+2x \implies h = -6# cm
This solution makes algebraically sense, since the area would be #(-5cm)(-6cm) = 30cm^2#, but we can't have sides of a rectangle with negative measures, so we must reject these solutions.
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Answer 2

#color(blue)("Length"=6cm)#

#color(blue)("Width"=5cm)#

The area of a rectangle is given by:

#"Area"="length"xx"width"#
We are given #"Area"=30cm^2#

Therefore:

#"Area"=(2x+4)*(5+2x)=30#
# \ \ \ \ \ \ \ \ \ \=4x^2+18x+20=30#
# \ \ \ \ \ \ \ \ \ \=4x^2+18x-10=0#

Factor:

#(2x+10)(2x-1)=0=>x=-5 and x=1/2#

We now check these, since we can't have a negative side to the rectangle:

#x=-5#
#(2(-5)+4)=-6#

We discard this solution.

#x=1/2#
#(2(1/2)+4)=5#
#(5+2(1/2))=6#

So:

#"Length"=6cm#
#"Width"=5cm#
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Answer 3

Length = (2x + 4) cm, Width = (5 + 2x) cm

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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