Solve 6 99x ^ 2 = (11x - 1) ^ 2 - 12 | 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

$$99 x ^ { 2 } = ( 11 x - 1 ) ^ { 2 } - 12$$

Solve for x

$x = \frac{\sqrt{3} + 1}{2} \approx 1.366025404$
$x=\frac{1-\sqrt{3}}{2}\approx -0.366025404$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(11x-1\right)^{2}$.

$$99x^{2}=121x^{2}-22x+1-12$$

Subtract $12$ from $1$ to get $-11$.

$$99x^{2}=121x^{2}-22x-11$$

Subtract $121x^{2}$ from both sides.

$$99x^{2}-121x^{2}=-22x-11$$

Combine $99x^{2}$ and $-121x^{2}$ to get $-22x^{2}$.

$$-22x^{2}=-22x-11$$

Add $22x$ to both sides.

$$-22x^{2}+22x=-11$$

Add $11$ to both sides.

$$-22x^{2}+22x+11=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-22$ for $a$, $22$ for $b$, and $11$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-22±\sqrt{22^{2}-4\left(-22\right)\times 11}}{2\left(-22\right)}$$

Square $22$.

$$x=\frac{-22±\sqrt{484-4\left(-22\right)\times 11}}{2\left(-22\right)}$$

Multiply $-4$ times $-22$.

$$x=\frac{-22±\sqrt{484+88\times 11}}{2\left(-22\right)}$$

Multiply $88$ times $11$.

$$x=\frac{-22±\sqrt{484+968}}{2\left(-22\right)}$$

Add $484$ to $968$.

$$x=\frac{-22±\sqrt{1452}}{2\left(-22\right)}$$

Take the square root of $1452$.

$$x=\frac{-22±22\sqrt{3}}{2\left(-22\right)}$$

Multiply $2$ times $-22$.

$$x=\frac{-22±22\sqrt{3}}{-44}$$

Now solve the equation $x=\frac{-22±22\sqrt{3}}{-44}$ when $±$ is plus. Add $-22$ to $22\sqrt{3}$.

$$x=\frac{22\sqrt{3}-22}{-44}$$

Divide $-22+22\sqrt{3}$ by $-44$.

$$x=\frac{1-\sqrt{3}}{2}$$

Now solve the equation $x=\frac{-22±22\sqrt{3}}{-44}$ when $±$ is minus. Subtract $22\sqrt{3}$ from $-22$.

$$x=\frac{-22\sqrt{3}-22}{-44}$$

Divide $-22-22\sqrt{3}$ by $-44$.

$$x=\frac{\sqrt{3}+1}{2}$$

The equation is now solved.

$$x=\frac{1-\sqrt{3}}{2}$$ $$x=\frac{\sqrt{3}+1}{2}$$

Steps for Completing the Square

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(11x-1\right)^{2}$.

$$99x^{2}=121x^{2}-22x+1-12$$

Subtract $12$ from $1$ to get $-11$.

$$99x^{2}=121x^{2}-22x-11$$

Subtract $121x^{2}$ from both sides.

$$99x^{2}-121x^{2}=-22x-11$$

Combine $99x^{2}$ and $-121x^{2}$ to get $-22x^{2}$.

$$-22x^{2}=-22x-11$$

Add $22x$ to both sides.

$$-22x^{2}+22x=-11$$

Divide both sides by $-22$.

$$\frac{-22x^{2}+22x}{-22}=-\frac{11}{-22}$$

Dividing by $-22$ undoes the multiplication by $-22$.

$$x^{2}+\frac{22}{-22}x=-\frac{11}{-22}$$

Divide $22$ by $-22$.

$$x^{2}-x=-\frac{11}{-22}$$

Reduce the fraction $\frac{-11}{-22}$ to lowest terms by extracting and canceling out $11$.

$$x^{2}-x=\frac{1}{2}$$

Divide $-1$, the coefficient of the $x$ term, by $2$ to get $-\frac{1}{2}$. Then add the square of $-\frac{1}{2}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square $-\frac{1}{2}$ by squaring both the numerator and the denominator of the fraction.

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

Add $\frac{1}{2}$ to $\frac{1}{4}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-x+\frac{1}{4}=\frac{3}{4}$$

Factor $x^{2}-x+\frac{1}{4}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{3}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{3}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{\sqrt{3}}{2}$$ $$x-\frac{1}{2}=-\frac{\sqrt{3}}{2}$$

Add $\frac{1}{2}$ to both sides of the equation.

$$x=\frac{\sqrt{3}+1}{2}$$ $$x=\frac{1-\sqrt{3}}{2}$$