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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The area of the shaded sector is one-fourth of the area of the whole circle so:
area of shaded region
The area of the shaded region is 9pi units^2
Step-by-step explanation:
y = 8x² + 8x - 96
y = 8(x² + x - 12)
y = 8(x + 4)(x - 3)
When y = 0, 8(x + 4)(x - 3) = 0.
=> x = -4 or x = 3. (B)
y=-3x+4
this is a line with slope -3 and y intercept of 4
the line is all the points that are the solutions to the equation
D.a line that shows the set of all solutions to the equation
