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Formula
Solve a quadratic inequality
Answer
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Graph
$\left ( x - 1 \right ) \left ( x - 3 \right ) > 0$
$\left ( x - 1 \right ) \left ( x - 3 \right ) > 0$
Solution of inequality
$x < 1 \text{ or } x > 3$
$\left( x-1 \right) \left( x-3 \right) > 0$
$x < 1 $ or $ x > 3$
$ $ Solve a solution to $ x$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) > 0$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } > 0$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } > 0$
$ $ Factorize the expression $ $
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) > 0$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) > \color{#FF6800}{ 0 }$
$ $ Values that can be $ \left ( x - 3 \right ) \left ( x - 1 \right ) > 0 $ are $ \begin{cases} x - 3 > 0 \\ x - 1 > 0 \end{cases} $ or $ \begin{cases} x - 3 < 0 \\ x - 1 < 0 \end{cases}$
$\begin{cases} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } > \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } > \color{#FF6800}{ 0 } \end{cases} \\ \begin{cases} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } < \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } < \color{#FF6800}{ 0 } \end{cases}$
$\begin{cases} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } > \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } > \color{#FF6800}{ 0 } \end{cases} \\ \begin{cases} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } < \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } < \color{#FF6800}{ 0 } \end{cases}$
$ $ Solve the inequality $ $
$\begin{cases} \color{#FF6800}{ x } > \color{#FF6800}{ 3 } \\ \color{#FF6800}{ x } > \color{#FF6800}{ 1 } \end{cases} \\ \begin{cases} \color{#FF6800}{ x } < \color{#FF6800}{ 3 } \\ \color{#FF6800}{ x } < \color{#FF6800}{ 1 } \end{cases}$
$\begin{cases} \color{#FF6800}{ x } > \color{#FF6800}{ 3 } \\ \color{#FF6800}{ x } > \color{#FF6800}{ 1 } \end{cases} \\ \begin{cases} x < 3 \\ x < 1 \end{cases}$
$ $ Find the intersection of sets of each interval $ $
$\color{#FF6800}{ x } > \color{#FF6800}{ 3 } \\ \begin{cases} x < 3 \\ x < 1 \end{cases}$
$x > 3 \\ \begin{cases} \color{#FF6800}{ x } < \color{#FF6800}{ 3 } \\ \color{#FF6800}{ x } < \color{#FF6800}{ 1 } \end{cases}$
$ $ Find the intersection of sets of each interval $ $
$x > 3 \\ \color{#FF6800}{ x } < \color{#FF6800}{ 1 }$
$\color{#FF6800}{ x } > \color{#FF6800}{ 3 } \\ \color{#FF6800}{ x } < \color{#FF6800}{ 1 }$
$ $ Find the union of sets of each interval $ $
$\color{#FF6800}{ x } < \color{#FF6800}{ 1 } $ or $ \color{#FF6800}{ x } > \color{#FF6800}{ 3 }$
$ $ 그래프 보기 $ $
Inequality
Solution search results
search-thumbnail-If the sum of two consecutive 
numbers is $45$ and one number is $X$ 
.This statement in the form of 
equation $1s:$ 
$\left(1$ Point) $\right)$ 
$○5x+1$ $1eft\left(x+1$ $r1gnt\right)=45s$ 
$○sx+1ef\left(x+2$ $r1gnt\right)=145s$ 
$sx+1x=45s$
7th-9th grade
Algebra
search-thumbnail-$s|ef\left(-1n$ $\left($ }\right)^{50}\ $\right)$ \ | | is\ equal\ to\ $S$ 
$s1S$ 
$S-1S$ 
$s2S$ 
$s50s$
7th-9th grade
Other
search-thumbnail-Given the set of ordered pairs $\left(\left(-7.0\right),\left(-6,5\right),\left(-5,-3\right),\left(-1,2\right)$ $\left(1,6\right),\left(2,-2\right)$ $\left(5,3\right)\left(7,-8\right)\right)$ 
Find f(7)fAleft(7\right) 
O a 
O b -8 
6. 
$5$
7th-9th grade
Algebra
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