Type a math problem

This site uses cookies for analytics, personalized content and ads. By continuing to browse this site, you agree to this use. Learn more

Type a math problem

Solve for x

x=2<br/>x=-1

$x=2$

$x=−1$

$x=−1$

Solution Steps

x-3+ \frac{ 4 }{ { x }^{ 2 } } =0

$x−3+x_{2}4 =0$

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x^{2}.

Variable $x$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $x_{2}$.

x^{2}x+x^{2}\left(-3\right)+4=0

$x_{2}x+x_{2}(−3)+4=0$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

To multiply powers of the same base, add their exponents. Add $2$ and $1$ to get $3$.

x^{3}+x^{2}\left(-3\right)+4=0

$x_{3}+x_{2}(−3)+4=0$

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

x^{3}-3x^{2}+4=0

$x_{3}−3x_{2}+4=0$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 4 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

By Rational Root Theorem, all rational roots of a polynomial are in the form $qp $, where $p$ divides the constant term $4$ and $q$ divides the leading coefficient $1$. List all candidates $qp $.

±4,±2,±1

$±4,±2,±1$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

x=-1

$x=−1$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-3x^{2}+4 by x+1 to get x^{2}-4x+4. Solve the equation where the result equals to 0.

By Factor theorem, $x−k$ is a factor of the polynomial for each root $k$. Divide $x_{3}−3x_{2}+4$ by $x+1$ to get $x_{2}−4x+4$. Solve the equation where the result equals to $0$.

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

$x_{2}−4x+4=0$

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, -4 for b, and 4 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 $a$, $−4$ for $b$, and $4$ for $c$ in the quadratic formula.

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

$x=2−(−4)±(−4)_{2}−4×1×4 $

Do the calculations.

Do the calculations.

x=\frac{4±0}{2}

$x=24±0 $

Solutions are the same.

Solutions are the same.

x=2

$x=2$

List all found solutions.

List all found solutions.

x=-1 x=2

$x=−1$ $x=2$

Graph

Graph Both Sides in 2D

Graph in 2D

Giving is as easy as 1, 2, 3

Get 1,000 points to donate to a school of your choice when you join Give With Bing

Share

Copy

Copied to clipboard

x^{2}x+x^{2}\left(-3\right)+4=0

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x^{2}.

x^{3}+x^{2}\left(-3\right)+4=0

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

x^{3}-3x^{2}+4=0

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

±4,±2,±1

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 4 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

x=-1

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

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

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-3x^{2}+4 by x+1 to get x^{2}-4x+4. Solve the equation where the result equals to 0.

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

x=\frac{4±0}{2}

Do the calculations.

x=2

Solutions are the same.

x=-1 x=2

List all found 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 $

Back to top