Type a math problem

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Type a math problem

Solve for x

x\in (-\infty,-5]\cup [3,\infty)

$x∈(−∞,−5]∪[3,∞)$

Steps Using the Quadratic Formula

x ^ { 2 } + 2 x - 15 \geq 0

$x_{2}+2x−15≥0$

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}+2x-15=0

$x_{2}+2x−15=0$

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, 2 for b, and -15 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 $a$, $2$ for $b$, and $−15$ for $c$ in the quadratic formula.

x=\frac{-2±\sqrt{2^{2}-4\times 1\left(-15\right)}}{2}

$x=2−2±2_{2}−4×1(−15) $

Do the calculations.

Do the calculations.

x=\frac{-2±8}{2}

$x=2−2±8 $

Solve the equation x=\frac{-2±8}{2} when ± is plus and when ± is minus.

Solve the equation $x=2−2±8 $ when $±$ is plus and when $±$ is minus.

x=3 x=-5

$x=3$ $x=−5$

Rewrite the inequality by using the obtained solutions.

Rewrite the inequality by using the obtained solutions.

\left(x-3\right)\left(x+5\right)\geq 0

$(x−3)(x+5)≥0$

For the product to be ≥0, x-3 and x+5 have to be both ≤0 or both ≥0. Consider the case when x-3 and x+5 are both ≤0.

For the product to be $≥0$, $x−3$ and $x+5$ have to be both $≤0$ or both $≥0$. Consider the case when $x−3$ and $x+5$ are both $≤0$.

x-3\leq 0 x+5\leq 0

$x−3≤0$ $x+5≤0$

The solution satisfying both inequalities is x\leq -5.

The solution satisfying both inequalities is $x≤−5$.

x\leq -5

$x≤−5$

Consider the case when x+5 and x-3 are both ≥0.

Consider the case when $x+5$ and $x−3$ are both $≥0$.

x+5\geq 0 x-3\geq 0

$x+5≥0$ $x−3≥0$

The solution satisfying both inequalities is x\geq 3.

The solution satisfying both inequalities is $x≥3$.

x\geq 3

$x≥3$

The final solution is the union of the obtained solutions.

The final solution is the union of the obtained solutions.

x\leq -5\text{; }x\geq 3

$x≤−5;x≥3$

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x^{2}+2x-15=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{-2±\sqrt{2^{2}-4\times 1\left(-15\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, 2 for b, and -15 for c in the quadratic formula.

x=\frac{-2±8}{2}

Do the calculations.

x=3 x=-5

Solve the equation x=\frac{-2±8}{2} when ± is plus and when ± is minus.

\left(x-3\right)\left(x+5\right)\geq 0

Rewrite the inequality by using the obtained solutions.

x-3\leq 0 x+5\leq 0

For the product to be ≥0, x-3 and x+5 have to be both ≤0 or both ≥0. Consider the case when x-3 and x+5 are both ≤0.

x\leq -5

The solution satisfying both inequalities is x\leq -5.

x+5\geq 0 x-3\geq 0

Consider the case when x+5 and x-3 are both ≥0.

x\geq 3

The solution satisfying both inequalities is x\geq 3.

x\leq -5\text{; }x\geq 3

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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