I need help on this math problem.
The width of a certain painting is 5cm less than twice the length. The length of the diagonal distance across the painting is 107cm. Find the length and width. Round your answer to 2 decimal places.
Can you check is my substitute is correct?
digonal distance: 107
width: 2L-5
length: L
a^2 + b^2 = c^2
(2L-5)^2 + L^2 = 107^2
please help and thank you
never mind found my other post!
Your substitution seems to be correct. To solve the equation and find the values of L and the width (W), we need to expand and simplify the equation:
(2L - 5)^2 + L^2 = 107^2
Expanding (2L - 5)^2:
4L^2 - 20L + 25 + L^2 = 107^2
Combine like terms:
5L^2 - 20L + 25 = 107^2
Now, subtract (107^2) from both sides to isolate the equation:
5L^2 - 20L + 25 - 107^2 = 0
Simplify further:
5L^2 - 20L + 25 - 11449 = 0
5L^2 - 20L - 11424 = 0
Now, we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. Let's use the quadratic formula to find the values of L:
L = (-b ± √(b^2 - 4ac)) / (2a)
For this equation, a = 5, b = -20, and c = -11424:
L = (-(-20) ± √((-20)^2 - 4(5)(-11424))) / (2(5))
L = (20 ± √(400 + 228480)) / 10
L = (20 ± √228880) / 10
Now, we can calculate the two possible values of L:
L1 = (20 + √228880) / 10
L2 = (20 - √228880) / 10
Compute the square root and simplify the values of L further. Once you have the values of L, substitute them back into the original equation to find the corresponding widths (2L - 5).