Solve ((x)^(2)-1)/(2)-(3x-1)/(4)=2 | 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

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

Solve for x

$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$
$x=3$

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring By Grouping

Multiply both sides of the equation by $4$, the least common multiple of $2,4$.

$$2\left(x^{2}-1\right)-\left(3x-1\right)=8$$

Use the distributive property to multiply $2$ by $x^{2}-1$.

$$2x^{2}-2-\left(3x-1\right)=8$$

To find the opposite of $3x-1$, find the opposite of each term.

$$2x^{2}-2-3x+1=8$$

Add $-2$ and $1$ to get $-1$.

$$2x^{2}-1-3x=8$$

Subtract $8$ from both sides.

$$2x^{2}-1-3x-8=0$$

Subtract $8$ from $-1$ to get $-9$.

$$2x^{2}-9-3x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$2x^{2}-3x-9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $2x^{2}+ax+bx-9$. To find $a$ and $b$, set up a system to be solved.

$$a+b=-3$$ $$ab=2\left(-9\right)=-18$$

Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product $-18$.

$$1,-18$$ $$2,-9$$ $$3,-6$$

Calculate the sum for each pair.

$$1-18=-17$$ $$2-9=-7$$ $$3-6=-3$$

The solution is the pair that gives sum $-3$.

$$a=-6$$ $$b=3$$

Rewrite $2x^{2}-3x-9$ as $\left(2x^{2}-6x\right)+\left(3x-9\right)$.

$$\left(2x^{2}-6x\right)+\left(3x-9\right)$$

Factor out $2x$ in the first and $3$ in the second group.

$$2x\left(x-3\right)+3\left(x-3\right)$$

Factor out common term $x-3$ by using distributive property.

$$\left(x-3\right)\left(2x+3\right)$$

To find equation solutions, solve $x-3=0$ and $2x+3=0$.

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

Steps Using the Quadratic Formula

Multiply both sides of the equation by $4$, the least common multiple of $2,4$.

$$2\left(x^{2}-1\right)-\left(3x-1\right)=8$$

Use the distributive property to multiply $2$ by $x^{2}-1$.

$$2x^{2}-2-\left(3x-1\right)=8$$

To find the opposite of $3x-1$, find the opposite of each term.

$$2x^{2}-2-3x+1=8$$

Add $-2$ and $1$ to get $-1$.

$$2x^{2}-1-3x=8$$

Subtract $8$ from both sides.

$$2x^{2}-1-3x-8=0$$

Subtract $8$ from $-1$ to get $-9$.

$$2x^{2}-9-3x=0$$

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.

$$2x^{2}-3x-9=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $2$ for $a$, $-3$ for $b$, and $-9$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 2\left(-9\right)}}{2\times 2}$$

Square $-3$.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 2\left(-9\right)}}{2\times 2}$$

Multiply $-4$ times $2$.

$$x=\frac{-\left(-3\right)±\sqrt{9-8\left(-9\right)}}{2\times 2}$$

Multiply $-8$ times $-9$.

$$x=\frac{-\left(-3\right)±\sqrt{9+72}}{2\times 2}$$

Add $9$ to $72$.

$$x=\frac{-\left(-3\right)±\sqrt{81}}{2\times 2}$$

Take the square root of $81$.

$$x=\frac{-\left(-3\right)±9}{2\times 2}$$

The opposite of $-3$ is $3$.

$$x=\frac{3±9}{2\times 2}$$

Multiply $2$ times $2$.

$$x=\frac{3±9}{4}$$

Now solve the equation $x=\frac{3±9}{4}$ when $±$ is plus. Add $3$ to $9$.

$$x=\frac{12}{4}$$

Divide $12$ by $4$.

$$x=3$$

Now solve the equation $x=\frac{3±9}{4}$ when $±$ is minus. Subtract $9$ from $3$.

$$x=-\frac{6}{4}$$

Reduce the fraction $\frac{-6}{4}$ to lowest terms by extracting and canceling out $2$.

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

The equation is now solved.

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

Steps for Completing the Square

Multiply both sides of the equation by $4$, the least common multiple of $2,4$.

$$2\left(x^{2}-1\right)-\left(3x-1\right)=8$$

Use the distributive property to multiply $2$ by $x^{2}-1$.

$$2x^{2}-2-\left(3x-1\right)=8$$

To find the opposite of $3x-1$, find the opposite of each term.

$$2x^{2}-2-3x+1=8$$

Add $-2$ and $1$ to get $-1$.

$$2x^{2}-1-3x=8$$

Add $1$ to both sides.

$$2x^{2}-3x=8+1$$

Add $8$ and $1$ to get $9$.

$$2x^{2}-3x=9$$

Divide both sides by $2$.

$$\frac{2x^{2}-3x}{2}=\frac{9}{2}$$

Dividing by $2$ undoes the multiplication by $2$.

$$x^{2}-\frac{3}{2}x=\frac{9}{2}$$

Divide $-\frac{3}{2}$, the coefficient of the $x$ term, by $2$ to get $-\frac{3}{4}$. Then add the square of $-\frac{3}{4}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{2}x+\left(-\frac{3}{4}\right)^{2}=\frac{9}{2}+\left(-\frac{3}{4}\right)^{2}$$

Square $-\frac{3}{4}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{9}{2}+\frac{9}{16}$$

Add $\frac{9}{2}$ to $\frac{9}{16}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{81}{16}$$

Factor $x^{2}-\frac{3}{2}x+\frac{9}{16}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{3}{4}\right)^{2}=\frac{81}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{81}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{9}{4}$$ $$x-\frac{3}{4}=-\frac{9}{4}$$

Add $\frac{3}{4}$ to both sides of the equation.

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