Solve 4x ^ 2 - 3x + 4 = 3x ^ 2 - 7x + 9 | 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

$$4x^{2}-3x+4=3x^{2}-7x+9$$

Solve for x

$x=-5$
$x=1$

Steps Using Factoring

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring

Subtract $3x^{2}$ from both sides.

$$4x^{2}-3x+4-3x^{2}=-7x+9$$

Combine $4x^{2}$ and $-3x^{2}$ to get $x^{2}$.

$$x^{2}-3x+4=-7x+9$$

Add $7x$ to both sides.

$$x^{2}-3x+4+7x=9$$

Combine $-3x$ and $7x$ to get $4x$.

$$x^{2}+4x+4=9$$

Subtract $9$ from both sides.

$$x^{2}+4x+4-9=0$$

Subtract $9$ from $4$ to get $-5$.

$$x^{2}+4x-5=0$$

To solve the equation, factor $x^{2}+4x-5$ using formula $x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right)$. To find $a$ and $b$, set up a system to be solved.

$$a+b=4$$ $$ab=-5$$

Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1$$ $$b=5$$

Rewrite factored expression $\left(x+a\right)\left(x+b\right)$ using the obtained values.

$$\left(x-1\right)\left(x+5\right)$$

To find equation solutions, solve $x-1=0$ and $x+5=0$.

$$x=1$$ $$x=-5$$

Steps Using Factoring By Grouping

Subtract $3x^{2}$ from both sides.

$$4x^{2}-3x+4-3x^{2}=-7x+9$$

Combine $4x^{2}$ and $-3x^{2}$ to get $x^{2}$.

$$x^{2}-3x+4=-7x+9$$

Add $7x$ to both sides.

$$x^{2}-3x+4+7x=9$$

Combine $-3x$ and $7x$ to get $4x$.

$$x^{2}+4x+4=9$$

Subtract $9$ from both sides.

$$x^{2}+4x+4-9=0$$

Subtract $9$ from $4$ to get $-5$.

$$x^{2}+4x-5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $x^{2}+ax+bx-5$. To find $a$ and $b$, set up a system to be solved.

$$a+b=4$$ $$ab=1\left(-5\right)=-5$$

Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1$$ $$b=5$$

Rewrite $x^{2}+4x-5$ as $\left(x^{2}-x\right)+\left(5x-5\right)$.

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

Factor out $x$ in the first and $5$ in the second group.

$$x\left(x-1\right)+5\left(x-1\right)$$

Factor out common term $x-1$ by using distributive property.

$$\left(x-1\right)\left(x+5\right)$$

To find equation solutions, solve $x-1=0$ and $x+5=0$.

$$x=1$$ $$x=-5$$

Steps Using the Quadratic Formula

Subtract $3x^{2}$ from both sides.

$$4x^{2}-3x+4-3x^{2}=-7x+9$$

Combine $4x^{2}$ and $-3x^{2}$ to get $x^{2}$.

$$x^{2}-3x+4=-7x+9$$

Add $7x$ to both sides.

$$x^{2}-3x+4+7x=9$$

Combine $-3x$ and $7x$ to get $4x$.

$$x^{2}+4x+4=9$$

Subtract $9$ from both sides.

$$x^{2}+4x+4-9=0$$

Subtract $9$ from $4$ to get $-5$.

$$x^{2}+4x-5=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $4$ for $b$, and $-5$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

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

Square $4$.

$$x=\frac{-4±\sqrt{16-4\left(-5\right)}}{2}$$

Multiply $-4$ times $-5$.

$$x=\frac{-4±\sqrt{16+20}}{2}$$

Add $16$ to $20$.

$$x=\frac{-4±\sqrt{36}}{2}$$

Take the square root of $36$.

$$x=\frac{-4±6}{2}$$

Now solve the equation $x=\frac{-4±6}{2}$ when $±$ is plus. Add $-4$ to $6$.

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

Divide $2$ by $2$.

$$x=1$$

Now solve the equation $x=\frac{-4±6}{2}$ when $±$ is minus. Subtract $6$ from $-4$.

$$x=-\frac{10}{2}$$

Divide $-10$ by $2$.

$$x=-5$$

The equation is now solved.

$$x=1$$ $$x=-5$$

Steps for Completing the Square

Subtract $3x^{2}$ from both sides.

$$4x^{2}-3x+4-3x^{2}=-7x+9$$

Combine $4x^{2}$ and $-3x^{2}$ to get $x^{2}$.

$$x^{2}-3x+4=-7x+9$$

Add $7x$ to both sides.

$$x^{2}-3x+4+7x=9$$

Combine $-3x$ and $7x$ to get $4x$.

$$x^{2}+4x+4=9$$

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

$$\left(x+2\right)^{2}=9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{9}$$

Simplify.

$$x+2=3$$ $$x+2=-3$$

Subtract $2$ from both sides of the equation.

$$x=1$$ $$x=-5$$