Module 3 Application Problem #5 You have 320 square feet of sod and you have two rectangular areas where you want to place sod. The sides of all the rectangles are in terms of x. For example, area A might be x+3 by x+10. Find the value or values of x that will work for the areas. Also, find all the lengths and widths of your sod rectangles. Be sure to use all the sod so there isn't any left over. At the end, you may round to the nearest tenth if needed. If you eliminate any solutions, justify why. Example: (x+3)(x+10)+(x+4)(x+7)=320 Step 1: Determine the additional feet for width (this can be any number, I would suggest keeping it below 10) Area 1 (X+3 ) Area 2 (X+6 ) Step 2: Determine the additional feet for length (this can be any number, but must be higher than number used for width) Area 1 (X+ 5 ) Area 2 (X+ 8 ) Step 3: Create equation to show total area of both rectangular portions needing sod. (X+3 ) (X+6 ) + (x+ 5 ) (X + 8 Step 4: the value of X? Work: A=(x+3)(x+6) A=(x+5)(x+8) X= 8 Total Area= (x+3)(x+6)+(x+5)(x+8)=320 x^2+ 6x+3x+18+x^2+8x+5x+40=320 ) =320 2x^2+ 17x+58=320 2x^2+ 17x-262-0 Solving for x x= 7.959 and -16.459 We discard the negative vatue because lengths and widths cannot be negative or non-real. Therefore, x=8 (rounded to the nearest tenth)
Module 3 Application Problem #5 You have 320 square feet of sod and you have two rectangular areas where you want to place sod. The sides of all the rectangles are in terms of x. For example, area A might be x+3 by x+10. Find the value or values of x that will work for the areas. Also, find all the lengths and widths of your sod rectangles. Be sure to use all the sod so there isn't any left over. At the end, you may round to the nearest tenth if needed. If you eliminate any solutions, justify why. Example: (x+3)(x+10)+(x+4)(x+7)=320 Step 1: Determine the additional feet for width (this can be any number, I would suggest keeping it below 10) Area 1 (X+3 ) Area 2 (X+6 ) Step 2: Determine the additional feet for length (this can be any number, but must be higher than number used for width) Area 1 (X+ 5 ) Area 2 (X+ 8 ) Step 3: Create equation to show total area of both rectangular portions needing sod. (X+3 ) (X+6 ) + (x+ 5 ) (X + 8 Step 4: the value of X? Work: A=(x+3)(x+6) A=(x+5)(x+8) X= 8 Total Area= (x+3)(x+6)+(x+5)(x+8)=320 x^2+ 6x+3x+18+x^2+8x+5x+40=320 ) =320 2x^2+ 17x+58=320 2x^2+ 17x-262-0 Solving for x x= 7.959 and -16.459 We discard the negative vatue because lengths and widths cannot be negative or non-real. Therefore, x=8 (rounded to the nearest tenth)
Module 3 Application Problem #5 You have 320 square feet of sod and you have two rectangular areas where you want to place sod. The sides of all the rectangles are in terms of x. For example, area A might be x+3 by x+10. Find the value or values of x that will work for the areas. Also, find all the lengths and widths of your sod rectangles. Be sure to use all the sod so there isn't any left over. At the end, you may round to the nearest tenth if needed. If you eliminate any solutions, justify why. Example: (x+3)(x+10)+(x+4)(x+7)=320 Step 1: Determine the additional feet for width (this can be any number, I would suggest keeping it below 10) Area 1 (X+3 ) Area 2 (X+6 ) Step 2: Determine the additional feet for length (this can be any number, but must be higher than number used for width) Area 1 (X+ 5 ) Area 2 (X+ 8 ) Step 3: Create equation to show total area of both rectangular portions needing sod. (X+3 ) (X+6 ) + (x+ 5 ) (X + 8 Step 4: the value of X? Work: A=(x+3)(x+6) A=(x+5)(x+8) X= 8 Total Area= (x+3)(x+6)+(x+5)(x+8)=320 x^2+ 6x+3x+18+x^2+8x+5x+40=320 ) =320 2x^2+ 17x+58=320 2x^2+ 17x-262-0 Solving for x x= 7.959 and -16.459 We discard the negative vatue because lengths and widths cannot be negative or non-real. Therefore, x=8 (rounded to the nearest tenth)
My teacher is asking me what was my work for solving for x once I got my quadratic equation to 0? I’m trying to figure out for the future how I would write that for it.
Transcribed Image Text:Module 3 Application Problem #5
You have 320 square feet of sod and you have two rectangular areas where you want to place sod.
The sides of all the rectangles are in terms of x. For example, area A might be x+3 by x+10. Find the
value or values of x that will work for the areas. Also, find all the lengths and widths of your sod
rectangles. Be sure to use all the sod so there isn't any left over. At the end, you may round to the
nearest tenth if needed. If you eliminate any solutions, justify why.
Example: (x+3)(x+10)+(x+4) (x+7)=320
Step 4: the value of X?
Step 1: Determine the additional feet for width (this can be any number, I would suggest keeping it below 10)
Area 1
(X+3
Area 2
(X+ 6
)
Step 2: Determine the additional feet for length (this can be any number, but must be higher than number used for
width)
Area 1
(X+ 5
)
Area 2 (X+ 8
Step 3: Create equation to show total area of both rectangular portions needing sod.
(X+3 ) (X+6) + (x+ 5
) (X + 8
Slide 6 of 12
Algebra 2, Part 1
Unit 1 Application Problems
Work:
A=(x+3)(x+6)
A=(x+5)(x+8)
X= 8
Total Area= (x+3)(x+6) +(x+5)(x+8)=320
x^2+ 6x+3x+18+x^2+8x+5x+40=320
2x^2+ 17x+58=320
2x^2+ 17x-262=0
Solving for x
x= 7.959 and -16.459
We discard the negative vatue because lengths and widths cannot be
negative or non-real.
Therefore, x=8 (rounded to the nearest tenth)
Notes
Comments
Mathematics
HB
AF
) =320
+ 92% 53
Formula Formula A polynomial with degree 2 is called a quadratic polynomial. A quadratic equation can be simplified to the standard form: ax² + bx + c = 0 Where, a ≠ 0. A, b, c are coefficients. c is also called "constant". 'x' is the unknown quantity
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