+1886 \(\frac{{x^2+x+3}}{{2x^2+x-6}}\ge 0?\)
Answer is:
\(x\in \big(-\infty ,-2\big) \cup \big( \frac{3}{2},+\infty \big)\)
+1886 \(\frac{{x^2+x+3}}{{2x^2+x-6}}\ge 0?\)
Answer is:
\(x\in \big(-\infty ,-2\big) \cup \big( \frac{3}{2},+\infty \big)\)
+1904 \(\frac{{x}^{2}+x+3}{{2x}^{2}+x-6}≥0\)
\(\frac{{x}^{2}+x+3}{{2x}^{2}+x-6}\times({2x}^{2}+x-6)≥0\times({2x}^{2}+x-6)\)
\(\frac{({x}^{2}+x+3)({2x}^{2}+x-6)}{{2x}^{2}+x-6}≥0\times({2x}^{2}+x-6)\)
\({x}^{2}+x+3≥0\times({2x}^{2}+x-6)\)
\({x}^{2}+x+3≥0\)
\(x ≥ {-1 \pm \sqrt{1^2-4(1)(3)} \over 2(1)}\)
\(x ≥ {-1 \pm \sqrt{1-4(1)(3)} \over 2(1)}\)
\(x ≥ {-1 \pm \sqrt{1-4(3)} \over 2(1)}\)
\(x ≥ {-1 \pm \sqrt{1-12} \over 2(1)}\)
\(x ≥ {-1 \pm \sqrt{-11} \over 2(1)}\)
\(x ≥ {-1 \pm \sqrt{11}i \over 2(1)}\)
\(x ≥ {-1 \pm \sqrt{11}i \over 2}\)
.
For the fraction to be positive, either top and bottom both need to be positive or both need to be negative.
The top line is always positive (so this also rules out the zero possibility).
Check this by either finding out whether or not
\(\displaystyle y=x^2+x+3\)
intersects with the x-axis, (by solving the equation with y = 0), or finding out where it's minimum point is.
The bottom line is positive for all values of x outside the range -2 to 3/2, (it's zero for those two values of x).
(As with the top line, it's an upward opening parabola).
(Solve the quadratic \(2x^{2}+x-6=0\) ).
So, the fraction is postive for all values of x less than -2 and for all values of x greater than 3/2.
Tiggsy.
