Given that the equation x ^ 2 + kx-3 = 0 and x ^ 2-4x - (k-1) = 0 have only one equal root, find the value of K and the same root | Math enthusiast | Mathematical art

Given that the equation x ^ 2 + kx-3 = 0 and x ^ 2-4x - (k-1) = 0 have only one equal root, find the value of K and the same root

Let the root of the equation x ^ 2 + kx-3 = 0 equal to x ^ 2-4x - (k-1) = 0 be a
Then a & # 178; + ka-3 = 0 and a & # 178; - 4A - (k-1) = 0
So a & # 178; + ka-3 = A & # 178; - 4A - (k-1)
(k+4)a=4-k
a=(4-k)/(4+k)
Substituting into the equation A & # 178; + ka-3 = 0, we have:
(4-k)²/(4+k)² +k(4-k)/(4+k)-3=0
(4-k)²+k(16-k²)-3(4+k)²=0
The results are as follows
k^3 +2k²+16k+32=0
That is, (K & # 178; + 16) (K + 2) = 0
The solution is k = - 2
So the equation is X & # 178; - 2x-3 = 0 and X & # 178; - 4x + 3 = 0
It is easy to know that the same root is x = 3

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