Let a, B and C be the lengths of three sides of the triangle ABC, and prove that x + 2aX + B = 0 and X + 2cx - B = 0 have common roots if and only if the angle a = 90 degree

Let a, B and C be the lengths of three sides of the triangle ABC, and prove that x + 2aX + B = 0 and X + 2cx - B = 0 have common roots if and only if the angle a = 90 degree

Let x + 2aX + B = x + 2cx-b get x = B ^ 2 / c-a. substitute x = B ^ 2 / C-A into x + 2aX + B or x + 2cx-b to get B ^ 2 + C ^ 2 = a ^ 2, so angle a = 90 degrees

It is known that the two roots of the equation x & # 178; - (2a-1) x + 4 (A-1) = 0 about X are the lengths of the two right angles of the right triangle with the length of the hypotenuse of 5, so s △ can be obtained

x1+x2=2a-1
x1x2=4a-4
x1²+x2²
=(x1+x2)²-2x1x2
=(2a-1)²-2(4a-4)
=4a²-4a+1-8a+8
=4a²-12a+9
=25
a²-3a-4=0
(a-4)(a+1)=0
a=4 b=3
S=1/2*4*3=6

It is known that the lengths of the two sides of a right triangle are the two roots of the equation x & # 178; - 16x + 55 = 0, and the lengths of the third side are obtained

The results show that (X-11) (X-5) = 0
Then x = 11 and x = 5
Then the third side length is the square root of (11 & # 178; + 5 & # 178;)
Equal to 146 under the root sign