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The sides of a triangle are 3/10, x+1/5 and 3/2x. What is the value of x?

 Dec 14, 2016
 #1
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I'm pretty sure you can find the answer to this using a graphing calculator if you set up the following

 

y1=.3

y2=x+.2

y3=1.5x

 

after that just find where they all intersect.

 

I am absolutely terrible at math so I would wait for someone else to confirm or deny that what I am doing is right.

 Dec 14, 2016
 #2
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I have assumed your triangle to be right-angle triangle. And also that 3/2x is the hypotenuse:

 (3/2x)^2 =(x+1/5)^2 + (0.3)^2

 

Solve for x:
(9 x^2)/4 = 0.09 + (x + 1/5)^2

0.09 + (x + 1/5)^2 = 9/100 + (x + 1/5)^2:
(9 x^2)/4 = 9/100 + (x + 1/5)^2

Expand out terms of the right hand side:
(9 x^2)/4 = x^2 + (2 x)/5 + 13/100

Subtract x^2 + (2 x)/5 + 13/100 from both sides:
(5 x^2)/4 - (2 x)/5 - 13/100 = 0

The left hand side factors into a product with three terms:
1/100 (5 x + 1) (25 x - 13) = 0

Multiply both sides by 100:
(5 x + 1) (25 x - 13) = 0

Split into two equations:
5 x + 1 = 0 or 25 x - 13 = 0

Subtract 1 from both sides:
5 x = -1 or 25 x - 13 = 0

Divide both sides by 5:
x = -1/5 or 25 x - 13 = 0

Add 13 to both sides:
x = -1/5 or 25 x = 13

Divide both sides by 25:
Answer: |x = -1/5        or        x = 13/25

 Dec 14, 2016

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