The shape of the room is not a square.
Pythagorean Theorem: 17^2+17^2= 578, square root 578 and you'll get 24.04.
The length of the diagonal of the floor of the room in the plan should have been 24.04 in order to be a square.
(x-8)(x+7)
so x2 makes the 2 x’s at the start then 8 times 7 is 56
Hello from MrBillDoesMath!
Answer: 6 u^2 + 4u + 6 = 0 where u = (x-5)^2
Discussion; I think the problem statement should actually be to rewrite <u>this</u> equation:
6 (x-5)^4 + 4 (x-5)^2 + 6 -= 0.
Note the power of "x-5" is 4 in the first term and is 2 in the second term. That is, the power of x-5 in the first terms is double, or the square, of the (x-5) power occurring in the second term. This suggest the substitution u = (x-5)^2. Then the equation can be rewritten as
6 u^2 + 4u + 6 = 0
which is a quadratic in "u".
Regards, MrB
