Type a math problem

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Type a math problem

Solve for x

x\in \left(-2,-1\right)

$x∈(−2,−1)$

Steps Using the Quadratic Formula

x ^ { 2 } + 3 x + 2 < 0

$x_{2}+3x+2<0$

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.

To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation $ax_{2}+bx+c=a(x−x_{1})(x−x_{2})$, where $x_{1}$ and $x_{2}$ are the solutions of the quadratic equation $ax_{2}+bx+c=0$.

x^{2}+3x+2=0

$x_{2}+3x+2=0$

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, 3 for b, and 2 for c in the quadratic formula.

All equations of the form $ax_{2}+bx+c=0$ can be solved using the quadratic formula: $2a−b±b_{2}−4ac $. Substitute $1$ for $a$, $3$ for $b$, and $2$ for $c$ in the quadratic formula.

x=\frac{-3±\sqrt{3^{2}-4\times 1\times 2}}{2}

$x=2−3±3_{2}−4×1×2 $

Do the calculations.

Do the calculations.

x=\frac{-3±1}{2}

$x=2−3±1 $

Solve the equation x=\frac{-3±1}{2} when ± is plus and when ± is minus.

Solve the equation $x=2−3±1 $ when $±$ is plus and when $±$ is minus.

x=-1 x=-2

$x=−1$ $x=−2$

Rewrite the inequality by using the obtained solutions.

Rewrite the inequality by using the obtained solutions.

\left(x+1\right)\left(x+2\right)<0

$(x+1)(x+2)<0$

For the product to be negative, x+1 and x+2 have to be of the opposite signs. Consider the case when x+1 is positive and x+2 is negative.

For the product to be negative, $x+1$ and $x+2$ have to be of the opposite signs. Consider the case when $x+1$ is positive and $x+2$ is negative.

x+1>0 x+2<0

$x+1>0$ $x+2<0$

This is false for any x.

This is false for any $x$.

x\in \emptyset

$x∈∅$

Consider the case when x+2 is positive and x+1 is negative.

Consider the case when $x+2$ is positive and $x+1$ is negative.

x+2>0 x+1<0

$x+2>0$ $x+1<0$

The solution satisfying both inequalities is x\in \left(-2,-1\right).

The solution satisfying both inequalities is $x∈(−2,−1)$.

x\in \left(-2,-1\right)

$x∈(−2,−1)$

The final solution is the union of the obtained solutions.

The final solution is the union of the obtained solutions.

x\in \left(-2,-1\right)

$x∈(−2,−1)$

Graph

Graph Inequality

Graph Both Sides in 2D

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x^{2}+3x+2=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{-3±\sqrt{3^{2}-4\times 1\times 2}}{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, 3 for b, and 2 for c in the quadratic formula.

x=\frac{-3±1}{2}

Do the calculations.

x=-1 x=-2

Solve the equation x=\frac{-3±1}{2} when ± is plus and when ± is minus.

\left(x+1\right)\left(x+2\right)<0

Rewrite the inequality by using the obtained solutions.

x+1>0 x+2<0

For the product to be negative, x+1 and x+2 have to be of the opposite signs. Consider the case when x+1 is positive and x+2 is negative.

x\in \emptyset

This is false for any x.

x+2>0 x+1<0

Consider the case when x+2 is positive and x+1 is negative.

x\in \left(-2,-1\right)

The solution satisfying both inequalities is x\in \left(-2,-1\right).

x\in \left(-2,-1\right)

The final solution is the union of the obtained solutions.

Examples

Quadratic equation

{ x } ^ { 2 } - 4 x - 5 = 0

$x_{2}−4x−5=0$

Trigonometry

4 \sin \theta \cos \theta = 2 \sin \theta

$4sinθcosθ=2sinθ$

Linear equation

y = 3x + 4

$y=3x+4$

Arithmetic

699 * 533

$699∗533$

Matrix

\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { - 1 } & { 1 } & { 5 } \end{array} \right]

$[25 34 ][2−1 01 35 ]$

Simultaneous equation

\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.

${8x+2y=467x+3y=47 $

Differentiation

\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }

$dxd (x−5)(3x_{2}−2) $

Integration

\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x

$∫_{0}xe_{−x_{2}}dx$

Limits

\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}

$x→−3lim x_{2}+2x−3x_{2}−9 $

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