3/x + x/2 < 0

Solving it by algebraic method leaves me with no real solution. But when I plotted a graph using the GC, my answer is x <0.

Can anyone explain?

Relevance
• 8 years ago

3/x + x/2 < 0

[2(3) / 2x] + (x² / 2x) < 0

(6 / 2x) + (x² / 2x) < 0

(6 + x²) / 2x < 0

If the quotient of two terms is negative, then one, and only one, term is negative.

If 6 + x² < 0,

x² < - 6

x < √(- 6)

x < √6 √(- 1)

x < √6 i

If 2x < 0,

x < 0

Since x < √6 i is imaginary, it does not exist, so

x < 0

¯¯¯¯

Source(s): 4/10/13
• Josh K
Lv 6
8 years ago

Multiply through by x

What you need to realize is that x can be negative in some circumstances affecting the inequality.

Case 1, x>0

3 + 1/2 x^2 < 0

x^2 < -6 Which is impossible in the real numbers, so this case yields no solution

Case 2, x = 0 undefined in original equation as a denominator = 0

Case 3 x < 0

3 + 1/2 x^2 > 0

1/2 x^2 > -3

x^2 > -6

Since x^2 will always be positive for any value of x, this will always be true.

Then, to come up with the solution, the criteria for this case is x<0 AND x any real number.

This intersection is identical to x < 0

So here, the defining characteristic of the case actually becomes the solution.

• 8 years ago

Anybody can see that all negative values of x is solutions for the inequality.

If we multiply both sides of the inequality by 2x we get (6+x^2)/2x < 0

for any negative number (6+x^2) is positive and 2x is negative, and this inequality is also fullfilled for all negative values of x.