Directions: Please understand the statement and answer correctly and completely.
Mr. (Name) cuts a rectangular plywod to be used for a furniture. The length of the plywood is twice its width and the area is 8^2 ft.
Questions:
1. What quadratic equation represents the area of a plywood? Write the equation in terms of the width of the plywood.
2. Solve the quadratic equation.
3. Which of the solutions or roots obtained represents the width of the plywood. Explain your answer.
4. What is the length of the plywood? Explain how you arrived at your answer.
Please answer what you can. Thank you.
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Answer:
Plywood1: x = width ; length = 2x
1) x (2x) = 4.5
2) 2x² - 4.5 = 0
a = 2 ; b = 0 ; c = -4.5
3) x = \frac{-b \pm\sqrt{b^2-4ac} }{2a} = \frac{0 \pm\sqrt{0-4(2)(-4.5)} }{2(2)}= \frac{\pm6}{4} = \boxed{\frac{3}{2}}x=
2a
−b±
b
2
−4ac
=
2(2)
0±
0−4(2)(−4.5)
=
4
±6
=
2
3
Plywood2: x = width ; length = 2x-1.4
1) x (2x-1.4) = 16
2) 2x² -1.4x -16 = 0
a = 2 ; b = -1.4 ; c = -16
3) x = \frac{-b \pm\sqrt{b^2-4ac} }{2a}=\frac{1.4 \pm\sqrt{1.96-4(2)(-16)} } < br / > {2(2)}=\frac{1.4 \pm11.4 }{4}=\boxed{3.2}x=
2a
−b±
b
2
−4ac
=
<
1.4±
1.96−4(2)(−16)
br/>2(2)=
4
1.4±11.4
=
3.2
Plywood3: x = width ; length = 5-x
1) x (5-x) = 6
2) -x² +5x -6 = 0
a = -1 ; b = 5 ; c = -6
3) x = \frac{-b \pm\sqrt{b^2-4ac} }{2a}=\frac{-5 \pm\sqrt{25-4(-1)(-6)} }{2(-1)}=\frac{-5 \pm1}{-2}x=
2a
−b±
b
2
−4ac
=
2(−1)
−5±
25−4(−1)(−6)
=
−2
−5±1
\boxed{x_1=3}
x
1
=3
\boxed{x_2=2}
x
2
=2
Answer:
Plywood1: x = width ; length = 2x
1) x (2x) = 4.5
2) 2x² - 4.5 = 0
a = 2 ; b = 0 ; c = -4.5
3) x = \frac{-b \pm\sqrt{b^2-4ac} }{2a} = \frac{0 \pm\sqrt{0-4(2)(-4.5)} }{2(2)}= \frac{\pm6}{4} = \boxed{\frac{3}{2}}x=
2a
−b±
b
2
−4ac
=
2(2)
0±
0−4(2)(−4.5)
=
4
±6
=
2
3
Plywood2: x = width ; length = 2x-1.4
1) x (2x-1.4) = 16
2) 2x² -1.4x -16 = 0
a = 2 ; b = -1.4 ; c = -16
3) x = \frac{-b \pm\sqrt{b^2-4ac} }{2a}=\frac{1.4 \pm\sqrt{1.96-4(2)(-16)} } < br / > {2(2)}=\frac{1.4 \pm11.4 }{4}=\boxed{3.2}x=
2a
−b±
b
2
−4ac
=
<
1.4±
1.96−4(2)(−16)
br/>2(2)=
4
1.4±11.4
=
3.2
Plywood3: x = width ; length = 5-x
1) x (5-x) = 6
2) -x² +5x -6 = 0
a = -1 ; b = 5 ; c = -6
3) x = \frac{-b \pm\sqrt{b^2-4ac} }{2a}=\frac{-5 \pm\sqrt{25-4(-1)(-6)} }{2(-1)}=\frac{-5 \pm1}{-2}x=
2a
−b±
b
2
−4ac
=
2(−1)
−5±
25−4(−1)(−6)
=
−2
−5±1
\boxed{x_1=3}
x
1
=3
\boxed{x_2=2}
x
2
=2