Solve for x
x\in \left(-\infty,\frac{-\sqrt{37}-5}{2}\right)\cup \left(\frac{\sqrt{37}-5}{2},\infty\right)
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x^{2}+5x-3=0
To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.
x=\frac{-5±\sqrt{5^{2}-4\times 1\left(-3\right)}}{2}
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, 5 for b, and -3 for c in the quadratic formula.
x=\frac{-5±\sqrt{37}}{2}
Do the calculations.
x=\frac{\sqrt{37}-5}{2} x=\frac{-\sqrt{37}-5}{2}
Solve the equation x=\frac{-5±\sqrt{37}}{2} when ± is plus and when ± is minus.
\left(x-\frac{\sqrt{37}-5}{2}\right)\left(x-\frac{-\sqrt{37}-5}{2}\right)>0
Rewrite the inequality by using the obtained solutions.
x-\frac{\sqrt{37}-5}{2}<0 x-\frac{-\sqrt{37}-5}{2}<0
For the product to be positive, x-\frac{\sqrt{37}-5}{2} and x-\frac{-\sqrt{37}-5}{2} have to be both negative or both positive. Consider the case when x-\frac{\sqrt{37}-5}{2} and x-\frac{-\sqrt{37}-5}{2} are both negative.
x<\frac{-\sqrt{37}-5}{2}
The solution satisfying both inequalities is x<\frac{-\sqrt{37}-5}{2}.
x-\frac{-\sqrt{37}-5}{2}>0 x-\frac{\sqrt{37}-5}{2}>0
Consider the case when x-\frac{\sqrt{37}-5}{2} and x-\frac{-\sqrt{37}-5}{2} are both positive.
x>\frac{\sqrt{37}-5}{2}
The solution satisfying both inequalities is x>\frac{\sqrt{37}-5}{2}.
x<\frac{-\sqrt{37}-5}{2}\text{; }x>\frac{\sqrt{37}-5}{2}
The final solution is the union of the obtained solutions.
Examples
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Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
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Limits
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