Two Algebra questions with quadratics!?

Question

#1-Two complex numbers a+bi and c+di are equal when a=c and b=d. Solve the equation 7x+5i=7-6yi for x andy, where x an y are real numbers. (I need to know how to do this by tomorrow)

#2- Use the quadratic formula to solve...

2x^2 + 7x + 6=0

Quadratic formula is x equals negative (B) plus or minus the square root of (b) squared minus four times (a) times (c) divided by two times (a).

Answer 1

1.

7x + 5i = 7 - 6yi

Isolate x:

7x = 7 - 6yi - 5i

Solve the right hand side in the proper form of complex numbers (a + bi)

7x = 7 + i(-6y - 5)
x = 1 + ( (-6y - 5)/7 ) i

Your goal is to get some value for a (the number without the i) and some value for b (the number next to the i). That's why I factored out the i so that we get the complex number proper form.

2.

The quadratic formula for ax^2 + bx + c = 0 goes as follows:

x = (-b +/- sqrt(b^2 - 4ac))/2a

In the above question, a = 2, b = 7, and c = 6. All you need to do is substitute. Note that "+/-" means "plus or minus", meaning you'll get two solutions for x.

Answer 2

1. x = 1; y = -5/6

Solution:

x = 7/7 = 1

y = -5i/6i = -5/6



2. x = -2 or -3/2

Solution:

b=7; a=2; c=6

Just plug in the values:

x = (-7 +- sqroot(7ˆ2 - 4*2*6))/2*2

x = (-7 +- sqroot(1))/4

2 solutions:

x = (-7-1)/4 = -8/4 = -2 or x = (-7+1)/4 = -6/4 = -3/2