Solve es the number that offered histor the number that offere number that 5x = 40x ^ (4/2) the indicial equation | 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

$$5 x = 40 x ^ { 2 }$$

Solve for x

$x=\frac{1}{8}=0.125$
$x=0$

Steps Using Factoring

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring

Subtract $40x^{2}$ from both sides.

$$5x-40x^{2}=0$$

Factor out $x$.

$$x\left(5-40x\right)=0$$

To find equation solutions, solve $x=0$ and $5-40x=0$.

$$x=0$$ $$x=\frac{1}{8}$$

Steps Using the Quadratic Formula

Subtract $40x^{2}$ from both sides.

$$5x-40x^{2}=0$$

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.

$$-40x^{2}+5x=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-40$ for $a$, $5$ for $b$, and $0$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-5±\sqrt{5^{2}}}{2\left(-40\right)}$$

Take the square root of $5^{2}$.

$$x=\frac{-5±5}{2\left(-40\right)}$$

Multiply $2$ times $-40$.

$$x=\frac{-5±5}{-80}$$

Now solve the equation $x=\frac{-5±5}{-80}$ when $±$ is plus. Add $-5$ to $5$.

$$x=\frac{0}{-80}$$

Divide $0$ by $-80$.

$$x=0$$

Now solve the equation $x=\frac{-5±5}{-80}$ when $±$ is minus. Subtract $5$ from $-5$.

$$x=-\frac{10}{-80}$$

Reduce the fraction $\frac{-10}{-80}$ to lowest terms by extracting and canceling out $10$.

$$x=\frac{1}{8}$$

The equation is now solved.

$$x=0$$ $$x=\frac{1}{8}$$

Steps for Completing the Square

Subtract $40x^{2}$ from both sides.

$$5x-40x^{2}=0$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form $x^{2}+bx=c$.

$$-40x^{2}+5x=0$$

Divide both sides by $-40$.

$$\frac{-40x^{2}+5x}{-40}=\frac{0}{-40}$$

Dividing by $-40$ undoes the multiplication by $-40$.

$$x^{2}+\frac{5}{-40}x=\frac{0}{-40}$$

Reduce the fraction $\frac{5}{-40}$ to lowest terms by extracting and canceling out $5$.

$$x^{2}-\frac{1}{8}x=\frac{0}{-40}$$

Divide $0$ by $-40$.

$$x^{2}-\frac{1}{8}x=0$$

Divide $-\frac{1}{8}$, the coefficient of the $x$ term, by $2$ to get $-\frac{1}{16}$. Then add the square of $-\frac{1}{16}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{8}x+\left(-\frac{1}{16}\right)^{2}=\left(-\frac{1}{16}\right)^{2}$$

Square $-\frac{1}{16}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{8}x+\frac{1}{256}=\frac{1}{256}$$

Factor $x^{2}-\frac{1}{8}x+\frac{1}{256}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{1}{16}\right)^{2}=\frac{1}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{16}\right)^{2}}=\sqrt{\frac{1}{256}}$$

Simplify.

$$x-\frac{1}{16}=\frac{1}{16}$$ $$x-\frac{1}{16}=-\frac{1}{16}$$

Add $\frac{1}{16}$ to both sides of the equation.

$$x=\frac{1}{8}$$ $$x=0$$