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A quadratic equation is a polynomial equation of second degree, meaning the highest exponent of its monomials is $2.$ A quadratic equation can be written in standard form as follows.

$ax_{2}+bx+c=0$

$Quadratic Equationax_{2}+bx+c=0 Related Functiony=ax_{2}+bx+c $

This type of equation can be solved using several methods, such as graphing and factoring.
A quadratic equation can be solved graphically by drawing the parabola that corresponds to the related quadratic function and identifying its zeros. Consider the following equation as an example.
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*expand_more* *exact* answer. To verify a solution, substitute it into $f(x)$ and evaluate the expression. Verifying the solution $x=-2$ is done by evaluating $f(-2).$
When $x=-2,$ the value of $f(x)$ is $0.$ Therefore, $x=-2$ is an exact solution. The solution $x=3$ can now be verified by following the same procedure.

Since $f(3)=0,$ $x=3$ is also a solution. Therefore, the equation $2x_{2}+3x−5=5x+7$ has two solutions.

$2x_{2}+3x−5=5x+7 $

These four steps can be followed to find the solutions of the equation.
1

Write the Equation in Standard Form

If neither side of the equation equals $0,$ rearrange it so that all terms are on the same side.

$2x_{2}+3x−5=5x+7⇕2x_{2}+3x−5−5x−7=0 $

If there is more than one term with the same degree, simplify the equation by combining like terms.
$2x_{2}+3x−5−5x−7=0⇕2x_{2}−2x−12=0 $

The equation is in standard form now. 2

Graph the Related Function

The side with the variables can now be seen as a function $f(x).$

$Quadratic Equation2x_{2}−2x−12=0Related Functionf(x)=2x_{2}−2x−12 $

To graph the quadratic function in standard form, its characteristics should be identified first. $f(x)=2x_{2}−2x−12$ | |||
---|---|---|---|

Direction | Vertex | Axis of Symmetry | $y-$intercept |

$a>0⇒$ upward | $(21 ,-225 )$ | $x=21 $ | $(0,-12)$ |

In addition to these points, the reflection of the $y-$intercept across the axis of symmetry is at $(1,-12),$ which is also on the parabola. Now, the graph can be drawn.

3

Identify Any Points With $y-$coordinate $0$

Now, find all the points on the graph that have a $y-$coordinate equal to $0.$

4

Identify the $x-$coordinates of Those Points

The $x-$coordinates of the identified points solve the equation $f(x)=0.$

In the example, the $x-$coordinates of the points where the curve intercepts the $x-$axis are $-2$ and $3.$ Note that solving an equation graphically does not necessarily lead to an$f(x)=2x_{2}−2x−12$

Substitute

$x=-2$

$f(-2)=2(-2)_{2}−2(-2)−12$

Evaluate right-hand side

CalcPow

Calculate power

$f(-2)=2(4)−2(-2)−12$

Multiply

Multiply

$f(-2)=8−2(-2)−12$

MultPosNeg

$a(-b)=-a⋅b$

$f(-2)=8−(-4)−12$

SubNeg

$a−(-b)=a+b$

$f(-2)=8+4−12$

AddSubTerms

Add and subtract terms

$f(-2)=0$

Solution | Substitute | Evaluate |
---|---|---|

$x=-2$ | $f(-2)=2(-2)_{2}−2(-2)−12$ | $f(-2)=0✓$ |

$x=3$ | $f(3)=2(3)_{2}−2(3)−12$ | $f(3)=0✓$ |

$x=-2andx=3 $

$ax_{2}+c=0$

and are solved using inverse operations. Once $x_{2}$ remains on the left-hand side, the equation can be written as $x_{2}=d,$
where $d=a-c $. The value of $d$ gives the number of solutions the equation has. $d>0d=0d<0 :2real solutions:1real solution:0real solutions $

Simple quadratic equations are quadratic equations whose linear coefficient $b$ is equal to $0.$
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*expand_more*

$ax_{2}+c=0 $

This type of equation can be solved using inverse operations, and two steps must be followed. - Isolate $x_{2}.$
- Take square roots on both sides of the equation.

1

Isolate $x_{2}$

2

Take Square Roots

Now that $x_{2}$ has been isolated, it is necessary to *undo* the exponent. Exponents and radicals of the same index undo each other. Therefore, square roots undo powers of $2.$
Note that the negative solution is also considered along with the principal root when solving the equation.

$x_{2}=100$

SqrtEqn

$LHS =RHS $

$x_{2} =100 $

$a_{2} =±a$

$x=±100 $

CalcRoot

Calculate root

$x=±10$

StateSol

State solutions

$x=10x=-10 $

When trying to factor a quadratic trinomial of the form $ax_{2}+bx+c,$ it can be difficult to *expand_more*
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seeits factors. Consider the following expression.

$8x_{2}+26x+6 $

Here, $a=8,$ $b=26,$ and $c=6.$ There are six steps to factor this trinomial.
1

Factor Out the GCF of $a,$ $b,$ and $c$

To fully factor a quadratic trinomial, the Greatest Common Factor (GCF) of $a,$ $b,$ and $c$ has to be factored out first. To identify the GCF of these numbers, their prime factors will be listed.
In the remaining steps, the factored coefficient $2$ before the parentheses can be ignored. The new considered quadratic trinomial is $4x_{2}+13x+3.$ Therefore, the current values of $a,$ $b,$ and $c,$ are $4,$ $13,$ and $3,$ respectively. If the GCF of the coefficients is $1,$ this step can be ignored.

$8=26=6= 2⋅2⋅22⋅132⋅3 $

It can be seen that $8,$ $26,$ and $6$ share exactly one factor, $2.$
$GCF(8,26,6)=2 $

Now, $2$ can be factored out.
$8x_{2}+26x+6$

SplitIntoFactors

Split into factors

$2(4)x_{2}+2(13)x+2(3)$

FactorOut

Factor out $2$

$2(4x_{2}+13x+3)$

2

Find the Factor Pair of $ac$ Whose Sum Is $b$

It is known that $a=4$ and $c=3,$ so $ac=12>0.$ Therefore, the factors must have the same sign. Also, $b=13.$ Since the sum of the factors is positive and they must have the same sign, both factors must be positive. All positive factor pairs of $12$ can now be listed and their sums checked.

Factors of $ac$ | Sum of Factors |
---|---|

$1$ and $12$ | $1+12=13✓$ |

$2$ and $6$ | $2+6=8×$ |

$3$ and $4$ | $3+4=7×$ |

In this case, the correct factor pair is $1$ and $12.$ The following table sums up how to determine the signs of the factors based on the values of $ac$ and $b.$

$ac$ | $b$ | Factors |
---|---|---|

Positive | Positive | Both positive |

Positive | Negative | Both negative |

Negative | Positive | One positive and one negative. The absolute value of the positive factor is greater. |

Negative | Negative | One positive and one negative. The absolute value of the negative factor is greater. |

Such analysis makes the list of possible factor pairs shorter.

3

Write $bx$ as a Sum

The factor pair obtained in the previous step will be used to rewrite the $x-$term — the linear term — of the quadratic trinomial as a sum. Remember that the factors are $1$ and $12.$
The linear term $13x$ can be rewritten in the original expression as $12x+x.$

$4x_{2}+13x+3⇕4x_{2}+12x+x+3 $

4

Factor Out the GCF of the First Two Terms

The expression has four terms, which can be grouped into the $first$ $two$ and the $last$ $two$ $terms.$ Then, the GCF of each group can be factored out.

$4x_{2}+12x+x+3 $

The first two terms, $4x_{2}$ and $12x,$ can be factored.
$4x_{2}=12x= 2⋅2⋅x⋅x2⋅2⋅3⋅x $

The GCF of $4x_{2}$ and $12x$ is $2⋅2⋅x=4x.$
5

Factor Out the GCF of the Last Two Terms

6

Factor Out the Common Factor

If all the previous steps have been performed correctly, there should now be two terms with a common factor.

$4x(x+3)+1(x+3) $

The common factor will be factored out.
The factored form of $4x_{2}+13x+3$ is $(4x+1)(x+3).$ Remember that the original trinomial was $8x_{2}+26x+6$ and that the GCF $2$ was factored out in Step $1.$ This GCF has to be included in the final result.
$8x_{2}+26x+6=2(4x+1)(x+3) $

$x_{2}−2x−3=0$

Show Solution *expand_more*

We'll focus on the left-hand side of the equation before solving for $x.$ Notice that $1$ and $-3$ are the factors that multiply to equal $-3$ and add to $-2.$
Notice that, in factored form, the numbers in the parentheses with $x$ are the numbers from the chosen factor pair. This is because the coefficient of $x_{2}$ is $1.$ We could have written the expression in factored form immediately after finding the factor pair. Lastly, we can use the zero product property to solve the equation.

$x_{2}−2x−3=0$

Rewrite

Rewrite $-2x$ as $x−3x$

$x_{2}+x−3x−3=0$

FactorOut

Factor out $x$

$x(x+1)−3x−3=0$

FactorOut

Factor out $-3$

$x(x+1)−3(x+1)=0$

FactorOut

Factor out $(x+1)$

$(x+1)(x−3)=0$

$x+1=0x−3=0 ↔x=-1↔x=-3 $

Thus, the solutions to the equation $x_{2}−2x−3=0$ are $x=-1$ and $x=3.$ {{ 'mldesktop-placeholder-grade' | message }} {{ article.displayTitle }}!

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