Solve নিচের তথ্যের আলোকে x - 1/x = 4; x ^ 3 - 1/(x ^ 3) এর মান কত ? | 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

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

Solve for x

$x=\sqrt{5}+2\approx 4.236067977$
$x=2-\sqrt{5}\approx -0.236067977$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

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

$$xx-1=4x$$

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

$$x^{2}-1=4x$$

Subtract $4x$ from both sides.

$$x^{2}-1-4x=0$$

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.

$$x^{2}-4x-1=0$$

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

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

Square $-4$.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-1\right)}}{2}$$

Multiply $-4$ times $-1$.

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

Add $16$ to $4$.

$$x=\frac{-\left(-4\right)±\sqrt{20}}{2}$$

Take the square root of $20$.

$$x=\frac{-\left(-4\right)±2\sqrt{5}}{2}$$

The opposite of $-4$ is $4$.

$$x=\frac{4±2\sqrt{5}}{2}$$

Now solve the equation $x=\frac{4±2\sqrt{5}}{2}$ when $±$ is plus. Add $4$ to $2\sqrt{5}$.

$$x=\frac{2\sqrt{5}+4}{2}$$

Divide $4+2\sqrt{5}$ by $2$.

$$x=\sqrt{5}+2$$

Now solve the equation $x=\frac{4±2\sqrt{5}}{2}$ when $±$ is minus. Subtract $2\sqrt{5}$ from $4$.

$$x=\frac{4-2\sqrt{5}}{2}$$

Divide $4-2\sqrt{5}$ by $2$.

$$x=2-\sqrt{5}$$

The equation is now solved.

$$x=\sqrt{5}+2$$ $$x=2-\sqrt{5}$$

Steps for Completing the Square

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

$$xx-1=4x$$

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

$$x^{2}-1=4x$$

Subtract $4x$ from both sides.

$$x^{2}-1-4x=0$$

Add $1$ to both sides. Anything plus zero gives itself.

$$x^{2}-4x=1$$

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

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

Square $-2$.

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

Add $1$ to $4$.

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

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

$$\left(x-2\right)^{2}=5$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{5}$$

Simplify.

$$x-2=\sqrt{5}$$ $$x-2=-\sqrt{5}$$

Add $2$ to both sides of the equation.

$$x=\sqrt{5}+2$$ $$x=2-\sqrt{5}$$