Solve Write the quadratic equation in standard form: - 11 - x ^ 2 = 5 - 5x ^ 2 - 2x | 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

$$\left. \begin{array} { l } { \quad - 11 - x ^ { 2 } = 5 - 5 x ^ { 2 } - 2 x } \end{array} \right.$$

Solve for x

$x = \frac{\sqrt{65} - 1}{4} \approx 1.765564437$
$x=\frac{-\sqrt{65}-1}{4}\approx -2.265564437$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Subtract $5$ from both sides.

$$-11-x^{2}-5=-5x^{2}-2x$$

Subtract $5$ from $-11$ to get $-16$.

$$-16-x^{2}=-5x^{2}-2x$$

Add $5x^{2}$ to both sides.

$$-16-x^{2}+5x^{2}=-2x$$

Combine $-x^{2}$ and $5x^{2}$ to get $4x^{2}$.

$$-16+4x^{2}=-2x$$

Add $2x$ to both sides.

$$-16+4x^{2}+2x=0$$

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.

$$4x^{2}+2x-16=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $4$ for $a$, $2$ for $b$, and $-16$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-2±\sqrt{2^{2}-4\times 4\left(-16\right)}}{2\times 4}$$

Square $2$.

$$x=\frac{-2±\sqrt{4-4\times 4\left(-16\right)}}{2\times 4}$$

Multiply $-4$ times $4$.

$$x=\frac{-2±\sqrt{4-16\left(-16\right)}}{2\times 4}$$

Multiply $-16$ times $-16$.

$$x=\frac{-2±\sqrt{4+256}}{2\times 4}$$

Add $4$ to $256$.

$$x=\frac{-2±\sqrt{260}}{2\times 4}$$

Take the square root of $260$.

$$x=\frac{-2±2\sqrt{65}}{2\times 4}$$

Multiply $2$ times $4$.

$$x=\frac{-2±2\sqrt{65}}{8}$$

Now solve the equation $x=\frac{-2±2\sqrt{65}}{8}$ when $±$ is plus. Add $-2$ to $2\sqrt{65}$.

$$x=\frac{2\sqrt{65}-2}{8}$$

Divide $-2+2\sqrt{65}$ by $8$.

$$x=\frac{\sqrt{65}-1}{4}$$

Now solve the equation $x=\frac{-2±2\sqrt{65}}{8}$ when $±$ is minus. Subtract $2\sqrt{65}$ from $-2$.

$$x=\frac{-2\sqrt{65}-2}{8}$$

Divide $-2-2\sqrt{65}$ by $8$.

$$x=\frac{-\sqrt{65}-1}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{65}-1}{4}$$ $$x=\frac{-\sqrt{65}-1}{4}$$

Steps for Completing the Square

Add $5x^{2}$ to both sides.

$$-11-x^{2}+5x^{2}=5-2x$$

Combine $-x^{2}$ and $5x^{2}$ to get $4x^{2}$.

$$-11+4x^{2}=5-2x$$

Add $2x$ to both sides.

$$-11+4x^{2}+2x=5$$

Add $11$ to both sides.

$$4x^{2}+2x=5+11$$

Add $5$ and $11$ to get $16$.

$$4x^{2}+2x=16$$

Divide both sides by $4$.

$$\frac{4x^{2}+2x}{4}=\frac{16}{4}$$

Dividing by $4$ undoes the multiplication by $4$.

$$x^{2}+\frac{2}{4}x=\frac{16}{4}$$

Reduce the fraction $\frac{2}{4}$ to lowest terms by extracting and canceling out $2$.

$$x^{2}+\frac{1}{2}x=\frac{16}{4}$$

Divide $16$ by $4$.

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

Divide $\frac{1}{2}$, the coefficient of the $x$ term, by $2$ to get $\frac{1}{4}$. Then add the square of $\frac{1}{4}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square $\frac{1}{4}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=4+\frac{1}{16}$$

Add $4$ to $\frac{1}{16}$.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{65}{16}$$

Factor $x^{2}+\frac{1}{2}x+\frac{1}{16}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{65}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{65}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{\sqrt{65}}{4}$$ $$x+\frac{1}{4}=-\frac{\sqrt{65}}{4}$$

Subtract $\frac{1}{4}$ from both sides of the equation.

$$x=\frac{\sqrt{65}-1}{4}$$ $$x=\frac{-\sqrt{65}-1}{4}$$