Solve edguadralic equalion mu rth; x ^ 2 - (7/80) * x + 1/40 = 0 | 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

$$edguadralic\ equalion\ \mu rth; x^{2}-(\frac{7}{80})x+\frac{1}{40}=0$$

Solve for x (complex solution)

$x=\frac{7+\sqrt{591}i}{160}\approx 0.04375+0.151940572i$
$x=\frac{-\sqrt{591}i+7}{160}\approx 0.04375-0.151940572i$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

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.

$$x^{2}-\frac{7}{80}x+\frac{1}{40}=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $-\frac{7}{80}$ for $b$, and $\frac{1}{40}$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-\frac{7}{80}\right)±\sqrt{\left(-\frac{7}{80}\right)^{2}-4\times \frac{1}{40}}}{2}$$

Square $-\frac{7}{80}$ by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{7}{80}\right)±\sqrt{\frac{49}{6400}-4\times \frac{1}{40}}}{2}$$

Multiply $-4$ times $\frac{1}{40}$.

$$x=\frac{-\left(-\frac{7}{80}\right)±\sqrt{\frac{49}{6400}-\frac{1}{10}}}{2}$$

Add $\frac{49}{6400}$ to $-\frac{1}{10}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{7}{80}\right)±\sqrt{-\frac{591}{6400}}}{2}$$

Take the square root of $-\frac{591}{6400}$.

$$x=\frac{-\left(-\frac{7}{80}\right)±\frac{\sqrt{591}i}{80}}{2}$$

The opposite of $-\frac{7}{80}$ is $\frac{7}{80}$.

$$x=\frac{\frac{7}{80}±\frac{\sqrt{591}i}{80}}{2}$$

Now solve the equation $x=\frac{\frac{7}{80}±\frac{\sqrt{591}i}{80}}{2}$ when $±$ is plus. Add $\frac{7}{80}$ to $\frac{i\sqrt{591}}{80}$.

$$x=\frac{7+\sqrt{591}i}{2\times 80}$$

Divide $\frac{7+i\sqrt{591}}{80}$ by $2$.

$$x=\frac{7+\sqrt{591}i}{160}$$

Now solve the equation $x=\frac{\frac{7}{80}±\frac{\sqrt{591}i}{80}}{2}$ when $±$ is minus. Subtract $\frac{i\sqrt{591}}{80}$ from $\frac{7}{80}$.

$$x=\frac{-\sqrt{591}i+7}{2\times 80}$$

Divide $\frac{7-i\sqrt{591}}{80}$ by $2$.

$$x=\frac{-\sqrt{591}i+7}{160}$$

The equation is now solved.

$$x=\frac{7+\sqrt{591}i}{160}$$ $$x=\frac{-\sqrt{591}i+7}{160}$$

Steps for Completing the Square

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 $x^{2}+bx=c$.

$$x^{2}-\frac{7}{80}x+\frac{1}{40}=0$$

Subtract $\frac{1}{40}$ from both sides of the equation.

$$x^{2}-\frac{7}{80}x+\frac{1}{40}-\frac{1}{40}=-\frac{1}{40}$$

Subtracting $\frac{1}{40}$ from itself leaves $0$.

$$x^{2}-\frac{7}{80}x=-\frac{1}{40}$$

Divide $-\frac{7}{80}$, the coefficient of the $x$ term, by $2$ to get $-\frac{7}{160}$. Then add the square of $-\frac{7}{160}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{80}x+\left(-\frac{7}{160}\right)^{2}=-\frac{1}{40}+\left(-\frac{7}{160}\right)^{2}$$

Square $-\frac{7}{160}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{80}x+\frac{49}{25600}=-\frac{1}{40}+\frac{49}{25600}$$

Add $-\frac{1}{40}$ to $\frac{49}{25600}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{80}x+\frac{49}{25600}=-\frac{591}{25600}$$

Factor $x^{2}-\frac{7}{80}x+\frac{49}{25600}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{7}{160}\right)^{2}=-\frac{591}{25600}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{160}\right)^{2}}=\sqrt{-\frac{591}{25600}}$$

Simplify.

$$x-\frac{7}{160}=\frac{\sqrt{591}i}{160}$$ $$x-\frac{7}{160}=-\frac{\sqrt{591}i}{160}$$

Add $\frac{7}{160}$ to both sides of the equation.

$$x=\frac{7+\sqrt{591}i}{160}$$ $$x=\frac{-\sqrt{591}i+7}{160}$$