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:
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<u>Answer-</u>
<em>The values of x and y are </em><em>13</em><em> and </em><em>24</em><em>, respectively.</em>
<u>Solution-</u>
From the attachment, l and m are parallel lines and
---------------------1
As, when a traversal line intersects two parallel lines, same-side interior angles are supplementary, or they add up to 180 degrees.
---------------------2
As, an exterior angle is equal to the sum of the opposite interior angles.
Adding equation 1 and 2, we get,
Putting the value of x in equation 1, we get
Mode would be sets that are the same
Are the numbers 6,5,3,1,24? or is it 2,4?
