Finding vertex, discriminant, and roots in polynomial equations

[mustang]

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).

 

[Chen]

Have you even tried solving these problems? Can you please show us your work?

 

[M98Ranger]

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.