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Why is this tricky? Any 10th grader who knows the quadratic formula can solve it.
Since 147 = 49x3, and the square root of 49 is 7, the result can be simplified further.
Divide and conquer. m=7/2+x and n=7/2-x. (7/2+x)(7/2-x)-49=-147/4-x^2. x=±i(7/2)√3, m=(7/2)(1±i√3) and n=(7/2)(1∓i√3)
m+n=7 mn=49 m=(7±Sqrt[147]i)/2=3.5±0.5Sqrt[147]i n=(7±Sqrt[147]i)/2=3.5±0.5Sqrt[147]i
At about 5:26 you accidentally flip the + into a - before the square root.
It’s in my head.
I thought this was impossible until you started using complex numbers. This should have been stated in the question. I was assuming real numbers.
we don't find this problem "tricky" enough. )) You need to change the title.
Why is this tricky? Any 10th grader who knows the quadratic formula can solve it.
Since 147 = 49x3, and the square root of 49 is 7, the result can be simplified further.
Divide and conquer. m=7/2+x and n=7/2-x. (7/2+x)(7/2-x)-49=-147/4-x^2. x=±i(7/2)√3, m=(7/2)(1±i√3) and n=(7/2)(1∓i√3)
m+n=7 mn=49 m=(7±Sqrt[147]i)/2=3.5±0.5Sqrt[147]i n=(7±Sqrt[147]i)/2=3.5±0.5Sqrt[147]i
At about 5:26 you accidentally flip the + into a - before the square root.
It’s in my head.
I thought this was impossible until you started using complex numbers. This should have been stated in the question. I was assuming real numbers.
we don't find this problem "tricky" enough. )) You need to change the title.