How to Solve Complex Algebra 2 Problems?

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SUMMARY

This discussion focuses on solving complex Algebra 2 problems, specifically addressing equations and polynomial functions. For problem 2, the coordinates of the vertex of the equation y=a(x-r1)(x+r2) are determined to be (r1, 0) and (r2, 0). Problem 4 involves calculating the discriminant using the formula b²-4ac, resulting in 4a(r1)(r2). Problem 14 is solved by isolating the square root and applying the quadratic formula, yielding x=14. In problem 22, the coefficients b and c are found to be -4 and 8, respectively, while problem 23 results in the polynomial P(x)=x³-x²-7x+6.

PREREQUISITES
  • Understanding of quadratic equations and their properties
  • Knowledge of polynomial functions and their roots
  • Familiarity with the quadratic formula
  • Ability to manipulate square root equations
NEXT STEPS
  • Study the properties of quadratic equations and their graphs
  • Learn about polynomial long division and synthetic division
  • Explore the use of the quadratic formula in various contexts
  • Investigate complex numbers and their applications in polynomial equations
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Students studying Algebra 2, educators teaching algebra concepts, and anyone looking to improve their problem-solving skills in polynomial equations and functions.

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Consider the equation y=a(x-r1)(x+r2) for problem 2 & 4.
Problem 2. State the coordinates of the vertex.
Problem 4. State the value of the discruminant.

Problem 14. Solve: sqrt(x-4) + 10 = sqrt(x+4)





Problem 22.
Find integers b and c such that the equation x^3+bx^2+cx-10=0 has -2+i as a root.

Problem 23. If P(x) is a cibic polynomial such that P(-3)=P(-1)=P(2)=0 and P(0)=6, find P(x).
 
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Have you even tried solving these problems? Can you please show us your work?
 



For problem 2, the coordinates of the vertex can be found by setting the derivative of the equation to zero and solving for x. This will give the x-coordinate of the vertex. To find the y-coordinate, substitute the x-coordinate into the original equation. The coordinates of the vertex are (r1, 0) and (r2, 0).

For problem 4, the discriminant can be found by using the formula b^2-4ac, where a=1, b=0, and c=a(r1)(r2). The discriminant for this equation is 4a(r1)(r2).

For problem 14, we can solve by isolating the square root term and then squaring both sides to eliminate the square root. This will give us a quadratic equation that can be solved using the quadratic formula. The solution for this problem is x=14.

For problem 22, we can use the fact that if -2+i is a root, then -2-i must also be a root. This means that (x-(-2+i))(x-(-2-i)) must be a factor of the polynomial. We can expand this and compare it to the given equation to find that b=-4 and c=8.

For problem 23, we can use the fact that if -3, -1, and 2 are roots, then (x+3)(x+1)(x-2) must be a factor of the polynomial. We can expand this and use the given point (0,6) to find that P(x)=x^3-x^2-7x+6.
 

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