document.write( "Question 448999: Given 3 points
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Algebra.Com's Answer #308884 by ewatrrr(24785)\"\" \"About 
You can put this solution on YOUR website!
 
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document.write( "Hi,
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document.write( "find the Quadratic(ax^2 + bx + c = 0), Given 3 points
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document.write( "f(3)= -11
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document.write( "f(1)= -3
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document.write( "f(-2)= -21
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document.write( " 9a + 3b + c = -11 
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document.write( "  a + b + c = -3
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document.write( " 4a  -2b + c = -21
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document.write( "  a = -2, b = 4, c = -5
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document.write( " -2x^2 +4x -5 = 0\r
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Solved by pluggable solver: Using Cramer's Rule to Solve Systems with 3 variables

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document.write( "  First let \"A=%28matrix%283%2C3%2C9%2C3%2C1%2C1%2C1%2C1%2C4%2C-2%2C1%29%29\". This is the matrix formed by the coefficients of the given system of equations.
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document.write( "  Take note that the right hand values of the system are \"-11\", \"-3\", and \"-21\" and they are highlighted here: 
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document.write( "  These values are important as they will be used to replace the columns of the matrix A.
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document.write( "  Now let's calculate the the determinant of the matrix A to get \"abs%28A%29=30\". To save space, I'm not showing the calculations for the determinant. However, if you need help with calculating the determinant of the matrix A, check out this solver.
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document.write( "  Notation note: \"abs%28A%29\" denotes the determinant of the matrix A.
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document.write( "  Now replace the first column of A (that corresponds to the variable 'x') with the values that form the right hand side of the system of equations. We will denote this new matrix \"A%5Bx%5D\" (since we're replacing the 'x' column so to speak).
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document.write( "  Now compute the determinant of \"A%5Bx%5D\" to get \"abs%28A%5Bx%5D%29=-60\". Again, as a space saver, I didn't include the calculations of the determinant. Check out this solver to see how to find this determinant.
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document.write( "  To find the first solution, simply divide the determinant of \"A%5Bx%5D\" by the determinant of \"A\" to get: \"x=%28abs%28A%5Bx%5D%29%29%2F%28abs%28A%29%29=%28-60%29%2F%2830%29=-2\"
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document.write( "  So the first solution is \"x=-2\"
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document.write( "  We'll follow the same basic idea to find the other two solutions. Let's reset by letting \"A=%28matrix%283%2C3%2C9%2C3%2C1%2C1%2C1%2C1%2C4%2C-2%2C1%29%29\" again (this is the coefficient matrix).
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document.write( "  Now replace the second column of A (that corresponds to the variable 'y') with the values that form the right hand side of the system of equations. We will denote this new matrix \"A%5By%5D\" (since we're replacing the 'y' column in a way).
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document.write( "  Now compute the determinant of \"A%5By%5D\" to get \"abs%28A%5By%5D%29=120\".
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document.write( "  To find the second solution, divide the determinant of \"A%5By%5D\" by the determinant of \"A\" to get: \"y=%28abs%28A%5By%5D%29%29%2F%28abs%28A%29%29=%28120%29%2F%2830%29=4\"
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document.write( "  So the second solution is \"y=4\"
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document.write( "  Let's reset again by letting \"A=%28matrix%283%2C3%2C9%2C3%2C1%2C1%2C1%2C1%2C4%2C-2%2C1%29%29\" which is the coefficient matrix.
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document.write( "  Replace the third column of A (that corresponds to the variable 'z') with the values that form the right hand side of the system of equations. We will denote this new matrix \"A%5Bz%5D\" 
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document.write( "  Now compute the determinant of \"A%5Bz%5D\" to get \"abs%28A%5Bz%5D%29=-150\".
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document.write( "  To find the third solution, divide the determinant of \"A%5Bz%5D\" by the determinant of \"A\" to get: \"z=%28abs%28A%5Bz%5D%29%29%2F%28abs%28A%29%29=%28-150%29%2F%2830%29=-5\"
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document.write( "  So the third solution is \"z=-5\"
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document.write( "  ====================================================================================
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document.write( "  Final Answer:
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document.write( "  So the three solutions are \"x=-2\", \"y=4\", and \"z=-5\" giving the ordered triple (-2, 4, -5)
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document.write( "  Note: there is a lot of work that is hidden in finding the determinants. Take a look at this 3x3 Determinant Solver to see how to get each determinant.
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