Solve (5500)/(21)=(44)/(21)\times(x)^(3) | Equatix - Math Solver

Fractions & Decimals

Ratios & Proportions

Basic Number Theory

Integers & Absolute Values

Factors & Multiples

Prime Factorization

Basic Equations & Inequalities

Coordinate Plane & Graphing

Linear Equations & Inequalities

Quadratic Equations

Polynomials & Factoring

Functions & Graphs

Systems of Equations

Trigonometric Functions

Trigonometric Identities & Equations

Right Triangle Trigonometry

Graphing Trigonometric Functions

Limits & Continuity

Derivatives & Applications

Integrals & Applications

Series & Sequences

Multivariable Calculus

Functions & Graphs

Polynomial & Rational Functions

Exponential & Logarithmic Functions

Sequences & Series

Descriptive Statistics

Normal Distribution

Hypothesis Testing

Regression Analysis

Linear Programming

Sets & Counting

Eigenvalues & Eigenvectors

Linear Transformations

Atomic Structure

Thermochemistry

Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$\frac{ 5500 }{ 21 } = \frac{ 44 }{ 21 } \times { x }^{ 3 }$$

Solve for x (complex solution)

$x=\frac{-5\sqrt{3}i-5}{2}\approx -2.5-4.330127019i$
$x=5$
$x=\frac{-5+5\sqrt{3}i}{2}\approx -2.5+4.330127019i$

Solution Steps

Multiply both sides by $\frac{21}{44}$, the reciprocal of $\frac{44}{21}$.

$$\frac{5500}{21}\times \frac{21}{44}=x^{3}$$

Multiply $\frac{5500}{21}$ and $\frac{21}{44}$ to get $125$.

$$125=x^{3}$$

Swap sides so that all variable terms are on the left hand side.

$$x^{3}=125$$

Subtract $125$ from both sides.

$$x^{3}-125=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $-125$ and $q$ divides the leading coefficient $1$. List all candidates $\frac{p}{q}$.

$$±125,±25,±5,±1$$

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.

$$x=5$$

By Factor theorem, $x-k$ is a factor of the polynomial for each root $k$. Divide $x^{3}-125$ by $x-5$ to get $x^{2}+5x+25$. Solve the equation where the result equals to $0$.

$$x^{2}+5x+25=0$$

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 $a$, $5$ for $b$, and $25$ for $c$ in the quadratic formula.

$$x=\frac{-5±\sqrt{5^{2}-4\times 1\times 25}}{2}$$

Do the calculations.

$$x=\frac{-5±\sqrt{-75}}{2}$$

Solve the equation $x^{2}+5x+25=0$ when $±$ is plus and when $±$ is minus.

$$x=\frac{-5i\sqrt{3}-5}{2}$$ $$x=\frac{-5+5i\sqrt{3}}{2}$$

List all found solutions.

$$x=5$$ $$x=\frac{-5i\sqrt{3}-5}{2}$$ $$x=\frac{-5+5i\sqrt{3}}{2}$$

Solve for x

$x=5$

Solution Steps

Multiply both sides by $\frac{21}{44}$, the reciprocal of $\frac{44}{21}$.

$$\frac{5500}{21}\times \frac{21}{44}=x^{3}$$

Multiply $\frac{5500}{21}$ and $\frac{21}{44}$ to get $125$.

$$125=x^{3}$$

Swap sides so that all variable terms are on the left hand side.

$$x^{3}=125$$

Subtract $125$ from both sides.

$$x^{3}-125=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $-125$ and $q$ divides the leading coefficient $1$. List all candidates $\frac{p}{q}$.

$$±125,±25,±5,±1$$

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.

$$x=5$$

By Factor theorem, $x-k$ is a factor of the polynomial for each root $k$. Divide $x^{3}-125$ by $x-5$ to get $x^{2}+5x+25$. Solve the equation where the result equals to $0$.

$$x^{2}+5x+25=0$$

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 $a$, $5$ for $b$, and $25$ for $c$ in the quadratic formula.

$$x=\frac{-5±\sqrt{5^{2}-4\times 1\times 25}}{2}$$

Do the calculations.

$$x=\frac{-5±\sqrt{-75}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$x\in \emptyset$$

List all found solutions.

$$x=5$$