Solve 02 (x - 3) ^ 2 + (x + 3) ^ 2 = (x - 3)(x + 3) + 30 | 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

$$( x - 3 ) ^ { 2 } + ( x + 3 ) ^ { 2 } = ( x - 3 ) ( x + 3 ) + 30$$

Solve for x

$x=\sqrt{3}\approx 1.732050808$
$x=-\sqrt{3}\approx -1.732050808$

Steps by Finding Square Root

Steps Using the Quadratic Formula

Steps by Finding Square Root

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-3\right)^{2}$.

$$x^{2}-6x+9+\left(x+3\right)^{2}=\left(x-3\right)\left(x+3\right)+30$$

Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(x+3\right)^{2}$.

$$x^{2}-6x+9+x^{2}+6x+9=\left(x-3\right)\left(x+3\right)+30$$

Combine $x^{2}$ and $x^{2}$ to get $2x^{2}$.

$$2x^{2}-6x+9+6x+9=\left(x-3\right)\left(x+3\right)+30$$

Combine $-6x$ and $6x$ to get $0$.

$$2x^{2}+9+9=\left(x-3\right)\left(x+3\right)+30$$

Add $9$ and $9$ to get $18$.

$$2x^{2}+18=\left(x-3\right)\left(x+3\right)+30$$

Consider $\left(x-3\right)\left(x+3\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$. Square $3$.

$$2x^{2}+18=x^{2}-9+30$$

Add $-9$ and $30$ to get $21$.

$$2x^{2}+18=x^{2}+21$$

Subtract $x^{2}$ from both sides.

$$2x^{2}+18-x^{2}=21$$

Combine $2x^{2}$ and $-x^{2}$ to get $x^{2}$.

$$x^{2}+18=21$$

Subtract $18$ from both sides.

$$x^{2}=21-18$$

Subtract $18$ from $21$ to get $3$.

$$x^{2}=3$$

Take the square root of both sides of the equation.

$$x=\sqrt{3}$$ $$x=-\sqrt{3}$$

Steps Using the Quadratic Formula

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-3\right)^{2}$.

$$x^{2}-6x+9+\left(x+3\right)^{2}=\left(x-3\right)\left(x+3\right)+30$$

Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(x+3\right)^{2}$.

$$x^{2}-6x+9+x^{2}+6x+9=\left(x-3\right)\left(x+3\right)+30$$

Combine $x^{2}$ and $x^{2}$ to get $2x^{2}$.

$$2x^{2}-6x+9+6x+9=\left(x-3\right)\left(x+3\right)+30$$

Combine $-6x$ and $6x$ to get $0$.

$$2x^{2}+9+9=\left(x-3\right)\left(x+3\right)+30$$

Add $9$ and $9$ to get $18$.

$$2x^{2}+18=\left(x-3\right)\left(x+3\right)+30$$

Consider $\left(x-3\right)\left(x+3\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$. Square $3$.

$$2x^{2}+18=x^{2}-9+30$$

Add $-9$ and $30$ to get $21$.

$$2x^{2}+18=x^{2}+21$$

Subtract $x^{2}$ from both sides.

$$2x^{2}+18-x^{2}=21$$

Combine $2x^{2}$ and $-x^{2}$ to get $x^{2}$.

$$x^{2}+18=21$$

Subtract $21$ from both sides.

$$x^{2}+18-21=0$$

Subtract $21$ from $18$ to get $-3$.

$$x^{2}-3=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $0$ for $b$, and $-3$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{0±\sqrt{0^{2}-4\left(-3\right)}}{2}$$

Square $0$.

$$x=\frac{0±\sqrt{-4\left(-3\right)}}{2}$$

Multiply $-4$ times $-3$.

$$x=\frac{0±\sqrt{12}}{2}$$

Take the square root of $12$.

$$x=\frac{0±2\sqrt{3}}{2}$$

Now solve the equation $x=\frac{0±2\sqrt{3}}{2}$ when $±$ is plus.

$$x=\sqrt{3}$$

Now solve the equation $x=\frac{0±2\sqrt{3}}{2}$ when $±$ is minus.

$$x=-\sqrt{3}$$

The equation is now solved.

$$x=\sqrt{3}$$ $$x=-\sqrt{3}$$