SOLUTION: The directions say: Find the dimensions of the rectangle or triangle that has the given area. A=1/2BH Number46 the are is 119 square feet The base is (X+6) and the height is (X+3

Question 189248This question is from textbook McDougal Littel Algebra 1
: The directions say: Find the dimensions of the rectangle or triangle that has the given area. A=1/2BH
Number46 the are is 119 square feet
The base is (X+6) and the height is (X+3).
I need help putting it the quadric formula.
THIS IS URGENT I WOULD LIKE FOR IT TO BE ANSWERED TONIGHT.
This question is from textbook McDougal Littel Algebra 1

Found 2 solutions by jonvaliente, checkley75:
Answer by jonvaliente(64) (Show Source): You can put this solution on YOUR website!
By substituting all the given facts into the formula, we get:
A=119
B=(x+6)
H=(x+3)
So,

Multiplying both sides by 2, (to elimiate the 1/2), we get:

Simplifying we get:

Subtracting 238 from both sides we get:

Factoring:

x=-20 and/or x=11
If x=-20
B = (x+6) = -20 + 6 = -14
H = (x+3) = -20 + 3 = -17
Since the base and height of the triangle cannot be negative then this is not a solution,
If x=11
B = (x+6) = 11 + 6 = 17
H = (x+3) = 11 + 3 = 14
Therefore the base is 17 and the height is 14
Answer by checkley75(3666) (Show Source): You can put this solution on YOUR website!
119=bh/2
119=(x+6)(x+3)/2
(x+6)(x+3)=119*2
x^2+9x+18=238
x^2+9x+18-238=0
x^2+9x-220=0
(x-11)(x+20)=0
x-11=0
x=11 ans.
Proof:
119=(11+6)(11+3)/2
119=17*14/2
119=238/2
119=119

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