Percy solved the equation x2 + 7x + 12 = 12. His work is shown below. Is Percy correct? Explain.
2 answers:
= > x² + 7x + 12 = 12
= > x² + ( 4 + 3 )x + 12 = 12
= > x² + 4x + 3x + 12 = 12
= > x( x + 4 ) + 3( x + 4 ) = 12
= > ( x + 4 ) ( x + 3 ) = 12
Percy did correct till this step. But by doing like this, Percy can't get the values of the variable x.
Percy should follow the following steps :
= > x² + 7x + 12 = 12
Add -12 on both sides,
= > x² + 7x + 12 - 12 = 12 - 12
= > x² + 7x = 0
= > x( x + 7 ) = 0
= > ( x = 0 ) or ( x + 7 = 0 )
= > ( x = 0 ) or ( x = - 7 )
Hence, required value(s) of x is 0 or -7
= > x² + ( 4 + 3 )x + 12 = 12
= > x² + 4x + 3x + 12 = 12
= > x( x + 4 ) + 3( x + 4 ) = 12
= > ( x + 4 ) ( x + 3 ) = 12
Percy did correct till this step. But by doing like this, Percy can't get the values of the variable x.
Percy should follow the following steps :
= > x² + 7x + 12 = 12
Add -12 on both sides,
= > x² + 7x + 12 - 12 = 12 - 12
= > x² + 7x = 0
= > x( x + 7 ) = 0
= > ( x = 0 ) or ( x + 7 = 0 )
= > ( x = 0 ) or ( x = - 7 )
Hence, required value(s) of x is 0 or -7
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Let
x-----> the length of rectangle
y-----> the width of rectangle
we know that
perimeter of rectangle=2*[x+y]
perimeter of rectangle=82 m
82=2*[x+y]---> divide by 2 both sides---> 41=x+y--> y=41-x---> equation 1
Area of rectangle=x*y
substitute equation 1 in the area formula
Area=x*[41-x]----> 41x-x²
using a graph tool
see the attached figure
the vertex is the point (20.5,420.25)
that means
for x=20.5 m ( length of rectangle)
the area is 420.25 m²
y=420.25/20.5----> 20.5 m
the dimensions are
20.5 m x 20.5 m------> is a square
the answer part 1)
<span>the dimensions of the rectangular with Maximum area is a square with length side 20.5 meters
</span>
Part 2)<span>b) Suppose 41 barriers each 2m long, are used instead. Can the same area be enclosed?
</span>divide the length side of the square by 2
so
20.5/2=10.25--------> 10 barriers
the dimensions are 10 barriers x 10 barriers
10 barriers=10*2---> 20 m
the area enclosed with barriers is =20*20----> 400 m²
400 m² < 420.25 m²
so
the answer Part 2) is
<span>the area enclosed by the barriers is less than the area enclosed by the rope
</span>
Part 3)<span>How much more area can be enclosed if the rope is used instead of the barriers
</span>
area using the rope=420.25 m²
area using the barriers=400 m²
420.25-400=20.25 m²
the answer part 3) is
20.25 m²
x-----> the length of rectangle
y-----> the width of rectangle
we know that
perimeter of rectangle=2*[x+y]
perimeter of rectangle=82 m
82=2*[x+y]---> divide by 2 both sides---> 41=x+y--> y=41-x---> equation 1
Area of rectangle=x*y
substitute equation 1 in the area formula
Area=x*[41-x]----> 41x-x²
using a graph tool
see the attached figure
the vertex is the point (20.5,420.25)
that means
for x=20.5 m ( length of rectangle)
the area is 420.25 m²
y=420.25/20.5----> 20.5 m
the dimensions are
20.5 m x 20.5 m------> is a square
the answer part 1)
<span>the dimensions of the rectangular with Maximum area is a square with length side 20.5 meters
</span>
Part 2)<span>b) Suppose 41 barriers each 2m long, are used instead. Can the same area be enclosed?
</span>divide the length side of the square by 2
so
20.5/2=10.25--------> 10 barriers
the dimensions are 10 barriers x 10 barriers
10 barriers=10*2---> 20 m
the area enclosed with barriers is =20*20----> 400 m²
400 m² < 420.25 m²
so
the answer Part 2) is
<span>the area enclosed by the barriers is less than the area enclosed by the rope
</span>
Part 3)<span>How much more area can be enclosed if the rope is used instead of the barriers
</span>
area using the rope=420.25 m²
area using the barriers=400 m²
420.25-400=20.25 m²
the answer part 3) is
20.25 m²