Question

Consider the equations 7x^2-2x+9=2x^2-5x+8 a)rewrite the equation in the form ax^2+bx+c=0 b)Describe the solutions of the equation as real or non-real
C) solve the equation over the set of complex numbers

Answer 1

For part a) start by subtracting 2x^2 from both sides, adding 5x to both sides and subtracting 8 from both sides. That makes the entire right side move to the left side.

Answer 2

You now have:\[7x^2-2x+9=2x^2-5x+8\]

Answer 3

You have x^2 terms on both sides of this equation, x terms on both sides, and constant terms on both sides. Consolidate all terms on the left side, in the format
\[ax^2 + bx+ c=0\]

Answer 4

You have:\[7x^2-2x+9=2x^2-5x+8\]

Answer 5

You could subtract 8 from both sides, for starters. That would eliminate the constant, 8, on the right side. Please finish this work.

Answer 6

So, 5x^2+3x+1=0?

Answer 7

If so for part b how do you know if the solutions are real or non-real? I also need help with part c

Answer 8

For b I got non real solutions because I solve the quadratic equation and got a negative so I couldn't square it. Now I just need help with part c

Answer 9

\(7x^2-2x+9=2x^2-5x+8 \)

a)
\(5x^2 + 3x + 1= 0\)

b)
Use the discriminant of the quadratic formula. It is the expression inside the square root. It tells you the nature of the solutions. If the discriminant is negative, the equation has complex solutions. If the discriminant is zero, the equation has one real solution (or more accurately, it has two equal real solutions). If the discriminant is positive, the equation has two different real solutions.

\(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Now that you have \(5x^2 + 3x + 1 = 0\), you know a = 5, b = 3, and c = 1.
Use those values in the expression: \(b^2 - 4ac\), which is the discriminant to find the nature of the roots.

This will answer part b).

c)
Use the quadratic formula with the values of a, b, and c you used in part B0, and find the two solutions.

Answer 10

You are correct with parts a) and b).