How do you solve this equation?

Question

f(x) = (2x^2 - 8) / (3x^2 + 9x - 30)

Answer 1

your equation is: f(x) = 0?

OK, it is equivalent to:

2x^2 - 8 = 0 AND 3x^2 + 9x -30 not = 0

2x^2 - 8 = 0 => x = (plus or minus) sqrt(8/2)
or x = 2 or x = -2

Check that 3x^2 +9x -30 not= 0
x= 2: 3 * 4 + 18 - 30 = 12 -12 = 0 => NOT ok
x = -2: 12 -18 -30 = -6 - 30 = -36 Not =0 =>OK

There is only one solution: x = -2

Answer 2

First of all, unlike the others, I'm going to assume that you want the general solution and not necessarily the solution for when f(x) = 0.

If we let y = f(x) and solve this for x in terms of y:

You could multiply both sides by the denominator, simplify things to a quadratic in the form of ax^2 + bx + c = 0, and use the quadratic formula. But it might be less messy if we first try to factor and (we hope) cancel some terms:

y = (2x^2 - 8) / (3x^2 + 9x - 30)
y = 2(x^2 - 4) / 3(x^2 + 3x - 10)
y = 2(x+2)(x-2) / 3(x^2 + 3x - 10)
y = 2(x+2)(x-2) / 3(x-2)(x+5)
y = 2(x+2) / 3(x+5)
3(x+5) * y = 2(x+2)
(3x + 15) * y = 2x + 4
3xy + 15y = 2x + 4
3xy - 2x = 4 - 15y
x(3y - 2) = 4 - 15y
x = (4 - 15y) / (3y - 2)
Where again, y = f(x)

Note that for f(x) = 0, you get x = 4/ (-2) = -2

Answer 3

f(x) = (2x^2 - 8) / (3x^2 + 9x - 30)

Numerator: 2(x^2-4) = 2(x-2)(x+2), suggests roots of +2 and -2.

Denominator is zero at x=2, and is non-zero at x=-2. So quotient is undefined at x=2.

Root is x=-2.

Answer 4

(2x^2-8) / (3x^2+9x-30)

2(x^2-4) / 3(x^2+3x-10)

2(x-2)(x+2) /3(x-2)(x+5)

the 2 (x-2) cancel out

2(x+2) / 3(x+5)

3(x+5)*y 2(x+2)
3x+15*y=2x+4
3xy+15y=2x+4
3xy-2x=4-15y
x(3y-2)=4-15y
x=4-15y/3y-2

y=f(X)
when f( x)=0 it becomes 4/-2= x=-2

Answer 5

simplify:

2(x^2-4) / 3(x^2 + 3x -10)

2(x+2)(x-2) / 3 (x+5)(x-2)

(x-2) terms in numerator and demonator cancel

2(x+2) / 3(x+5)

if you want f(x) = 0, then x= -2

Answer 6

By studying, asking questions in class and after school tutoring.