Solve x ^ 2 + (1/(3x)) ^ 2 = 30 ^ 2 | 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

$$x^{2}+(\frac{1}{3x})^{2}=30^{2}$$

Solve for x

$x=\frac{\sqrt{8106}-\sqrt{8094}}{6}\approx 0.011111112$
$x=\frac{\sqrt{8094}-\sqrt{8106}}{6}\approx -0.011111112$
$x=\frac{-\sqrt{8094}-\sqrt{8106}}{6}\approx -29.999997942$
$x = \frac{\sqrt{8094} + \sqrt{8106}}{6} \approx 29.999997942$

Steps Using the Quadratic Formula

To raise $\frac{1}{3x}$ to a power, raise both numerator and denominator to the power and then divide.

$$x^{2}+\frac{1^{2}}{\left(3x\right)^{2}}=30^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply $x^{2}$ times $\frac{\left(3x\right)^{2}}{\left(3x\right)^{2}}$.

$$\frac{x^{2}\times \left(3x\right)^{2}}{\left(3x\right)^{2}}+\frac{1^{2}}{\left(3x\right)^{2}}=30^{2}$$

Since $\frac{x^{2}\times \left(3x\right)^{2}}{\left(3x\right)^{2}}$ and $\frac{1^{2}}{\left(3x\right)^{2}}$ have the same denominator, add them by adding their numerators.

$$\frac{x^{2}\times \left(3x\right)^{2}+1^{2}}{\left(3x\right)^{2}}=30^{2}$$

Calculate $30$ to the power of $2$ and get $900$.

$$\frac{x^{2}\times \left(3x\right)^{2}+1^{2}}{\left(3x\right)^{2}}=900$$

Expand $\left(3x\right)^{2}$.

$$\frac{x^{2}\times 3^{2}x^{2}+1^{2}}{\left(3x\right)^{2}}=900$$

Calculate $3$ to the power of $2$ and get $9$.

$$\frac{x^{2}\times 9x^{2}+1^{2}}{\left(3x\right)^{2}}=900$$

To multiply powers of the same base, add their exponents. Add $2$ and $2$ to get $4$.

$$\frac{x^{4}\times 9+1^{2}}{\left(3x\right)^{2}}=900$$

Calculate $1$ to the power of $2$ and get $1$.

$$\frac{x^{4}\times 9+1}{\left(3x\right)^{2}}=900$$

Expand $\left(3x\right)^{2}$.

$$\frac{x^{4}\times 9+1}{3^{2}x^{2}}=900$$

Calculate $3$ to the power of $2$ and get $9$.

$$\frac{x^{4}\times 9+1}{9x^{2}}=900$$

Subtract $900$ from both sides.

$$\frac{x^{4}\times 9+1}{9x^{2}}-900=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply $900$ times $\frac{9x^{2}}{9x^{2}}$.

$$\frac{x^{4}\times 9+1}{9x^{2}}-\frac{900\times 9x^{2}}{9x^{2}}=0$$

Since $\frac{x^{4}\times 9+1}{9x^{2}}$ and $\frac{900\times 9x^{2}}{9x^{2}}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{x^{4}\times 9+1-900\times 9x^{2}}{9x^{2}}=0$$

Do the multiplications in $x^{4}\times 9+1-900\times 9x^{2}$.

$$\frac{x^{4}\times 9+1-8100x^{2}}{9x^{2}}=0$$

Factor the expressions that are not already factored in $\frac{x^{4}\times 9+1-8100x^{2}}{9x^{2}}$.

$$\frac{9\left(x^{2}-\left(-\frac{1}{3}\sqrt{1822499}+450\right)\right)\left(x^{2}-\left(\frac{1}{3}\sqrt{1822499}+450\right)\right)}{9x^{2}}=0$$

Cancel out $9$ in both numerator and denominator.

$$\frac{\left(x^{2}-\left(-\frac{1}{3}\sqrt{1822499}+450\right)\right)\left(x^{2}-\left(\frac{1}{3}\sqrt{1822499}+450\right)\right)}{x^{2}}=0$$

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

$$\left(x^{2}-\left(-\frac{1}{3}\sqrt{1822499}+450\right)\right)\left(x^{2}-\left(\frac{1}{3}\sqrt{1822499}+450\right)\right)=0$$

To find the opposite of $-\frac{1}{3}\sqrt{1822499}+450$, find the opposite of each term.

$$\left(x^{2}+\frac{1}{3}\sqrt{1822499}-450\right)\left(x^{2}-\left(\frac{1}{3}\sqrt{1822499}+450\right)\right)=0$$

To find the opposite of $\frac{1}{3}\sqrt{1822499}+450$, find the opposite of each term.

$$\left(x^{2}+\frac{1}{3}\sqrt{1822499}-450\right)\left(x^{2}-\frac{1}{3}\sqrt{1822499}-450\right)=0$$

Use the distributive property to multiply $x^{2}+\frac{1}{3}\sqrt{1822499}-450$ by $x^{2}-\frac{1}{3}\sqrt{1822499}-450$ and combine like terms.

$$x^{4}-900x^{2}-\frac{1}{9}\left(\sqrt{1822499}\right)^{2}+202500=0$$

The square of $\sqrt{1822499}$ is $1822499$.

$$x^{4}-900x^{2}-\frac{1}{9}\times 1822499+202500=0$$

Multiply $-\frac{1}{9}$ and $1822499$ to get $-\frac{1822499}{9}$.

$$x^{4}-900x^{2}-\frac{1822499}{9}+202500=0$$

Add $-\frac{1822499}{9}$ and $202500$ to get $\frac{1}{9}$.

$$x^{4}-900x^{2}+\frac{1}{9}=0$$

Substitute $t$ for $x^{2}$.

$$t^{2}-900t+\frac{1}{9}=0$$

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 $a$, $-900$ for $b$, and $\frac{1}{9}$ for $c$ in the quadratic formula.

$$t=\frac{-\left(-900\right)±\sqrt{\left(-900\right)^{2}-4\times 1\times \frac{1}{9}}}{2}$$

Do the calculations.

$$t=\frac{900±\frac{2}{3}\sqrt{1822499}}{2}$$

Solve the equation $t=\frac{900±\frac{2}{3}\sqrt{1822499}}{2}$ when $±$ is plus and when $±$ is minus.

$$t=\frac{\sqrt{1822499}}{3}+450$$ $$t=-\frac{\sqrt{1822499}}{3}+450$$

Since $x=t^{2}$, the solutions are obtained by evaluating $x=±\sqrt{t}$ for each $t$.

$$x=\frac{\sqrt{8094}+\sqrt{8106}}{6}$$ $$x=-\frac{\sqrt{8094}+\sqrt{8106}}{6}$$ $$x=-\frac{\sqrt{8094}-\sqrt{8106}}{6}$$ $$x=\frac{\sqrt{8094}-\sqrt{8106}}{6}$$