Respuesta :
I'll just factor the above equation.
x² + 18x + 80
x² ⇒ x * x
80
can be:
1 x 80
2 x 40
4 x 20
5 x 16
8 x 10 Correct pair
(x+8)(x+10)
x(x+10) +8(x+10) ⇒ x² + 10x + 8x + 80 = x² + 18x + 80
x+8 = 0
x = -8
x+10 = 0
x = -10
x = -8
(-8)² + 18(-8) + 80 = 0
64 - 144 + 80 = 0
144 - 144 = 0
0 = 0
(-10)² + 18(-10) + 80 = 0
100 - 180 + 80 = 0
180 - 180 = 0
0 = 0
I think the algebra tiles will not be a good tool to use to factor the quadratic equation because the equation is not a perfect square quadratic equation.
x² + 18x + 80
x² ⇒ x * x
80
can be:
1 x 80
2 x 40
4 x 20
5 x 16
8 x 10 Correct pair
(x+8)(x+10)
x(x+10) +8(x+10) ⇒ x² + 10x + 8x + 80 = x² + 18x + 80
x+8 = 0
x = -8
x+10 = 0
x = -10
x = -8
(-8)² + 18(-8) + 80 = 0
64 - 144 + 80 = 0
144 - 144 = 0
0 = 0
(-10)² + 18(-10) + 80 = 0
100 - 180 + 80 = 0
180 - 180 = 0
0 = 0
I think the algebra tiles will not be a good tool to use to factor the quadratic equation because the equation is not a perfect square quadratic equation.
the correct question in the attached figure
We proceed to analyze each case
case a) [tex] n^{2}+ 3n + 40 = (n- 8)(n- 5) [/tex]
Expand the right side and compare with the left side
so
[tex] (n- 8)(n- 5)=n^{2} -5n-8n+40\\ (n- 8)(n- 5)=n^{2} -13n+40
[/tex]
[tex] n^{2}+ 3n + 40 [/tex] is not equal to [tex] n^{2} -13n+40 [/tex]
therefore
case a) is not the equation represented by ms. wilson’s model
case b) [tex] n^{2} + 13n + 40 = (n + 8)(n + 5) [/tex]
Expand the right side and compare with the left side
so
[tex] (n+ 8)(n+ 5)=n^{2} +5n+8n+40\\ (n+ 8)(n+ 5)=n^{2} +13n+40
[/tex]
[tex] n^{2} + 13n + 40 [/tex] is equal to [tex] n^{2} + 13n + 40 [/tex]
therefore
case b) is the equation represented by ms. wilson’s model
case c) [tex] n^{2} + 40n + 13 = (n + 8)(n + 5) [/tex]
Expand the right side and compare with the left side
so
[tex] (n+ 8)(n+ 5)=n^{2} +5n+8n+40\\ (n+ 8)(n+ 5)=n^{2} +13n+40
[/tex]
[tex] n^{2} + 40n + 13 [/tex] is not equal to [tex] n^{2} + 13n + 40 [/tex]
therefore
case c) is not the equation represented by ms. wilson’s model
case d) [tex] n^{2} + 40n + 3 = (n - 8)(n - 5) [/tex]
Expand the right side and compare with the left side
so
[tex] (n- 8)(n- 5)=n^{2} -5n-8n+40\\ (n- 8)(n- 5)=n^{2} -13n+40
[/tex]
[tex] n^{2} + 40n + 3 [/tex] is not equal to [tex] n^{2} - 13n + 40 [/tex]
therefore
case d) is not the equation represented by ms. wilson’s model
therefore
the answer is the case b) [tex] n^{2} + 13n + 40 = (n + 8)(n + 5) [/tex]