Math question?

Question

For which of the following values of b will there be 2 distinct real solutions to the equation 2x^2-bx+6 = 0 ?
A) 4 times radical 3
B) -4 times radical 3
C)-2
D)0
E)7

I could only choose one answer....so it would be E right?

Answer 1

The quadratic formula is:

..... -b ± sqrt(b² - 4ac)
x = ------------------------
.................. 2a

The number of roots can be determined from the sqrt value (called the discriminant).

All you have to do is test b² - 4ac and see if is positive or not.

Case 1: Discriminant is negative. Here you will have two *imaginary* roots.
Case 2: Discriminant is zero. Here you will have *one* real root.
Case 3: Discriminant is positive. Here you will have two real roots.

You want case 3 so solve:
b² - 4ac > 0

Your values are:
a = 2
b = (-b)
c = 6

So plug them in and solve the inequality:
(-b)² - 4(2)(6) > 0
(-b)² - 24 > 0
(-b)² > 24
|-b| > sqrt(24)
|b| > 4.89897949...

b < -4.89897949 or b > 4.89897949

So testing your answers:
A) 4√3 = sqrt(48) which is bigger than sqrt(24). This would have 2 real roots.
B) -4√3 = -sqrt(48) which is smaller than -sqrt(24). So this too would have 2 real roots.
C) -2 is neither bigger nor smaller, so it would result in 2 *imaginary* roots.
D) 0 is also in the middle, so it would result in 2 *imaginary* roots.
E) 7 is greater than 4√3 so it would result in 2 real roots.

The answers are:A, B and E

Answer 2

As I'm sure you know, the quadratic formula for finding the solutions of
Ax^2 + Bx + C = 0 is

x = (-B + or - sqrt(B^2 - 4AC))/(2A)

For your equation, A=2, B=-b, and C=6.

In order for the roots to be real, the term inside the sqrt sign, called the discriminant, must be nonnegative. However, if the discriminant is zero, you get a repeated root. Therefore, choose the value of b from the given list for which (-b)^2 - 4(2)(6) > 0.