Suppose we wish to determine whether or not two given polynomials with complex coefficients have a common root. Given two first-degree polynomials a0 + a1x and b0 + b1x, we seek a single value of x such that
Solving each of these equations for x we get x = -a0/a1 and x = -b0/b1 respectively, so in order for both equations to be satisfied simultaneously we must have a0/a1 = b0/b1, which can also be written as a0b1 - a1b0 = 0. Formally we can regard this system as two linear equations in the two quantities x0 and x1, and write them in matrix form as
Hence a non-trivial solution requires the vanishing of the determinant of the coefficient matrix, which again gives a0b1 - a1b0 = 0.
Now consider two polynomials of degree 2. In this case we seek a single value of x such that
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Answer:
This is a true statement.
Step-by-step explanation:
In order to find the line of symmetry, use the symmetry formula.
-b/2a
The variable a would be the coefficient of x^2 and b would be the coefficient of x.
-(4)/2(1)
-4/2
-2
4(2k - 3) + 1 = 8k - 11
8k - 12 + 1 = 8k - 11
8k - 11 = 8k - 11
So, these two equations equal to the same thing.
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Well all I can say is that the answer is 623
