Could anyone help me with a maths problem?

Question

If f(x) = (3x + a), determine the values of a fro which, f(x² + x + 2) - f(3x - a) = 0 will have really roots.

Answer 1

f(x) = 3x + a

 

f(x²+x+2) = 3(x²+x+2) + a = 3x² + 3x + (6+a)

 

f(3x-1) = 3(3x-a) + a = 9x - 2a

 

f(x²+x+2) - f(3x-1)

= 3x² + 3x + 6 + a - 9x + 2a

= 3x² - 6x + 6 + 3a

 

The roots are the values of x for which f(x) = 0:

 

0 = 3x² - 6x + 6 + 3a

 

We use the quadratic formula to solve for x:

 

x = (-b/2a) ± (1/2a)√(b²-4ac)

 

Where a = 3, b = -6 and c = 6+3a.  The roots will be real when the discriminant, the expression under the radical (b²-4ac), is ≥ 0:

 

(b²-4ac) ≥ 0

(-6)²-4(3)(6+3a) ≥ 0

36 -12(6+3a) ≥ 0

-12(6+3a) ≥ 36

6 + 3a ≤ -3           (Divided by -12, so we had the flip the inequality sign)

3a ≤ -9

a ≤ -3

 

The roots of f(x) will be real when a ≤ -3.