Type a math problem

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Type a math problem

Solve for x

x\in \left(3-\sqrt{2},\sqrt{2}+3\right)

$x∈(3−2 ,2 +3)$

Steps Using the Quadratic Formula

- 2 x ^ { 2 } + 12 x - 14 > 0

$−2x_{2}+12x−14>0$

Multiply the inequality by -1 to make the coefficient of the highest power in -2x^{2}+12x-14 positive. Since -1 is <0, the inequality direction is changed.

Multiply the inequality by -1 to make the coefficient of the highest power in $−2x_{2}+12x−14$ positive. Since $−1$ is $<0$, the inequality direction is changed.

2x^{2}-12x+14<0

$2x_{2}−12x+14<0$

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 $ax_{2}+bx+c=0$.

2x^{2}-12x+14=0

$2x_{2}−12x+14=0$

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 2 for a, -12 for b, and 14 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 $2$ for $a$, $−12$ for $b$, and $14$ for $c$ in the quadratic formula.

x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 2\times 14}}{2\times 2}

$x=2×2−(−12)±(−12)_{2}−4×2×14 $

Do the calculations.

Do the calculations.

x=\frac{12±4\sqrt{2}}{4}

$x=412±42 $

Solve the equation x=\frac{12±4\sqrt{2}}{4} when ± is plus and when ± is minus.

Solve the equation $x=412±42 $ when $±$ is plus and when $±$ is minus.

x=\sqrt{2}+3 x=3-\sqrt{2}

$x=2 +3$ $x=3−2 $

Rewrite the inequality by using the obtained solutions.

Rewrite the inequality by using the obtained solutions.

2\left(x-\left(\sqrt{2}+3\right)\right)\left(x-\left(3-\sqrt{2}\right)\right)<0

$2(x−(2 +3))(x−(3−2 ))<0$

For the product to be negative, x-\left(\sqrt{2}+3\right) and x-\left(3-\sqrt{2}\right) have to be of the opposite signs. Consider the case when x-\left(\sqrt{2}+3\right) is positive and x-\left(3-\sqrt{2}\right) is negative.

For the product to be negative, $x−(2 +3)$ and $x−(3−2 )$ have to be of the opposite signs. Consider the case when $x−(2 +3)$ is positive and $x−(3−2 )$ is negative.

x-\left(\sqrt{2}+3\right)>0 x-\left(3-\sqrt{2}\right)<0

$x−(2 +3)>0$ $x−(3−2 )<0$

This is false for any x.

This is false for any $x$.

x\in \emptyset

$x∈∅$

Consider the case when x-\left(3-\sqrt{2}\right) is positive and x-\left(\sqrt{2}+3\right) is negative.

Consider the case when $x−(3−2 )$ is positive and $x−(2 +3)$ is negative.

x-\left(3-\sqrt{2}\right)>0 x-\left(\sqrt{2}+3\right)<0

$x−(3−2 )>0$ $x−(2 +3)<0$

The solution satisfying both inequalities is x\in \left(3-\sqrt{2},\sqrt{2}+3\right).

The solution satisfying both inequalities is $x∈(3−2 ,2 +3)$.

x\in \left(3-\sqrt{2},\sqrt{2}+3\right)

$x∈(3−2 ,2 +3)$

The final solution is the union of the obtained solutions.

The final solution is the union of the obtained solutions.

x\in \left(3-\sqrt{2},\sqrt{2}+3\right)

$x∈(3−2 ,2 +3)$

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2x^{2}-12x+14<0

Multiply the inequality by -1 to make the coefficient of the highest power in -2x^{2}+12x-14 positive. Since -1 is <0, the inequality direction is changed.

2x^{2}-12x+14=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{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 2\times 14}}{2\times 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 2 for a, -12 for b, and 14 for c in the quadratic formula.

x=\frac{12±4\sqrt{2}}{4}

Do the calculations.

x=\sqrt{2}+3 x=3-\sqrt{2}

Solve the equation x=\frac{12±4\sqrt{2}}{4} when ± is plus and when ± is minus.

2\left(x-\left(\sqrt{2}+3\right)\right)\left(x-\left(3-\sqrt{2}\right)\right)<0

Rewrite the inequality by using the obtained solutions.

x-\left(\sqrt{2}+3\right)>0 x-\left(3-\sqrt{2}\right)<0

For the product to be negative, x-\left(\sqrt{2}+3\right) and x-\left(3-\sqrt{2}\right) have to be of the opposite signs. Consider the case when x-\left(\sqrt{2}+3\right) is positive and x-\left(3-\sqrt{2}\right) is negative.

x\in \emptyset

This is false for any x.

x-\left(3-\sqrt{2}\right)>0 x-\left(\sqrt{2}+3\right)<0

Consider the case when x-\left(3-\sqrt{2}\right) is positive and x-\left(\sqrt{2}+3\right) is negative.

x\in \left(3-\sqrt{2},\sqrt{2}+3\right)

The solution satisfying both inequalities is x\in \left(3-\sqrt{2},\sqrt{2}+3\right).

x\in \left(3-\sqrt{2},\sqrt{2}+3\right)

The final solution is the union of the obtained solutions.

Examples

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

$x_{2}−4x−5=0$

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

$4sinθcosθ=2sinθ$

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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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