Which Equation Can Be Used to Solve for B
Which Equation Can Be Used to Solve for B
two.6: Quadratic Equations
 Page ID
 84916
 Solve quadratic equations by factoring.
 Solve quadratic equations past the square root property.
 Solve quadratic equations by completing the square.
 Solve quadratic equations by using the quadratic formula.
The calculator monitor on the left in Figure \(\PageIndex{one}\) is a \(23.6\)inch model and the one on the right is a \(27\)inch model. Proportionally, the monitors appear very similar. If there is a limited corporeality of space and we desire the largest monitor possible, how do we determine which ane to choose? In this section, nosotros will learn how to solve problems such as this using four dissimilar methods.
Solving Quadratic Equations by Factoring
An equation containing a seconddegree polynomial is chosen a
quadratic equation. For example, equations such as \(2x^2 +3x−1=0\) and \(ten^2−iv= 0\) are quadratic equations. They are used in countless ways in the fields of engineering, architecture, finance, biological scientific discipline, and, of course, mathematics.
Ofttimes the easiest method of solving a quadratic equation is
factoring. Factoring means finding expressions that tin exist multiplied together to give the expression on 1 side of the equation.
If a quadratic equation tin can be factored, it is written as a product of linear terms. Solving by factoring depends on the zeroproduct property, which states that if \(a⋅b=0\), and so \(a = 0\) or \(b =0\), where a and b are real numbers or algebraic expressions. In other words, if the production of two numbers or 2 expressions equals zero, then one of the numbers or ane of the expressions must equal zero because zero multiplied by anything equals nada.
Multiplying the factors expands the equation to a string of terms separated past plus or minus signs. And then, in that sense, the operation of multiplication undoes the functioning of factoring. For example, aggrandize the factored expression \((ten−2)(ten+three)\) by multiplying the two factors together.
\[\begin{align*} (x2)(x+3)&= x^2+3x2x6\\ &= x^two+ten6\\ \finish{align*}\]
The product is a quadratic expression. Gear up equal to zero, \(x^ii+10−6= 0\) is a quadratic equation. If we were to factor the equation, we would get dorsum the factors nosotros multiplied.
The process of factoring a quadratic equation depends on the leading coefficient, whether it is \(1\) or another integer. We will look at both situations; but first, nosotros want to confirm that the equation is written in standard grade, \(ax^ii+bx+c=0\), where \(a\), \(b\), and \(c\) are real numbers, and \(a≠0\). The equation \(ten^2 +x−6= 0\) is in standard grade.
We can use the zeroproduction property to solve quadratic equations in which we beginning have to factor out the greatest mutual factor(GCF), and for equations that take special factoring formulas as well, such equally the difference of squares, both of which we will meet later in this department.
The
zeroproduct holding
states
If \(a⋅b=0\), then \(a=0\) or \(b=0\),
where \(a\) and \(b\) are real numbers or algebraic expressions.
A quadratic equation is an equation containing a seconddegree polynomial; for example
\[ax^2+bx+c=0\]
where \(a\), \(b\), and \(c\) are existent numbers, and if \(a≠0\), it is in standard form.
Solving Quadratics with a Leading Coefficient of \(1\)
In the quadratic equation \(x^ii +ten−6=0\), the leading coefficient, or the coefficient of \(x^2\), is \(one\). We have one method of factoring quadratic equations in this form.
 Notice 2 numbers whose product equals \(c\) and whose sum equals \(b\).
 Apply those numbers to write 2 factors of the form \((ten+k)\) or \((ten−k)\), where k is ane of the numbers found in stride one. Utilize the numbers exactly as they are. In other words, if the two numbers are \(1\) and \(−2\), the factors are \((x+ane)(x−2)\).
 Solve using the zeroproduct property past setting each factor equal to aught and solving for the variable.
Factor and solve the equation: \(x^ii+x−6=0\).
Solution
To factor \(x^2 +ten−six=0\), nosotros look for two numbers whose production equals \(−half dozen\) and whose sum equals \(1\). Begin past looking at the possible factors of \(−vi\).
\[one⋅(−6) \nonumber \]
\[(−6)⋅1 \nonumber \]
\[2⋅(−three) \nonumber \]
\[3⋅(−2) \nonumber \]
The last pair, \(iii⋅(−two)\) sums to \(1\), so these are the numbers. Note that merely one pair of numbers volition work. Then, write the factors.
\[(x−two)(x+three)=0 \nonumber \]
To solve this equation, nosotros use the cypherproduction holding. Set each factor equal to zero and solve.
\[\begin{align*} (tentwo)(ten+iii)&= 0\\ (10two)&= 0\\ x&= 2\\ (10+iii)&= 0\\ x&= 3 \finish{marshal*}\]
The two solutions are \(2\) and \(−3\). Nosotros can see how the solutions relate to the graph in Figure \(\PageIndex{2}\). The solutions are the xintercepts of \(x^two +10−6=0\).
Factor and solve the quadratic equation: \(x^ii−5x−six=0\).
 Answer

\((10−halfdozen)(x+one)=0\), \(x=6\), \(10=−one\)
Solve the quadratic equation by factoring: \(x^two+8x+15=0\).
Solution
Find two numbers whose product equals \(15\) and whose sum equals \(8\). List the factors of \(15\).
\[one⋅fifteen \nonumber \]
\[iii⋅v \nonumber \]
\[(−one)⋅(−15) \nonumber \]
\[(−3)⋅(−v) \nonumber \]
The numbers that add to \(8\) are \(3\) and \(5\). Then, write the factors, set up each factor equal to nil, and solve.
\[\begin{align*} (x+3)(10+five)&= 0\\ (10+3)&= 0\\ x&= 3\\ (x+5)&= 0\\ x&= 5 \end{align*}\]
The solutions are \(−3\) and \(−v\).
Solve the quadratic equation by factoring: \(10^ii−4x−21=0\).
 Answer

\((10−7)(x+3)=0\), \(10=7\), \(x=−3\)
Solve the departure of squares equation using the zeroproduct property: \(x^2−nine=0\).
Solution
Recognizing that the equation represents the departure of squares, nosotros can write the two factors by taking the square root of each term, using a minus sign equally the operator in one factor and a plus sign every bit the operator in the other. Solve using the zerofactor property.
\[\begin{align*} x^2nine&= 0\\ (ten3)(x+3)&= 0\\ tenthree&= 0\\ ten&= 3\\ (x+3)&= 0\\ ten&= iii \end{align*}\]
The solutions are \(3\) and \(−3\).
Solve by factoring: \(10^ii−25=0\).
 Answer

\((10+5)(10−v)=0, x=−v, x=v\)
Factoring and Solving a Quadratic Equation of Higher Guild
When the leading coefficient is not \(one\), we factor a quadratic equation using the method called
grouping, which requires four terms.
With the equation in standard course, let’s review the grouping procedures
 With the quadratic in standard form, \(ax^two+bx+c=0\), multiply \(a⋅c\).
 Discover ii numbers whose production equals ac and whose sum equals \(b\).
 Rewrite the equation replacing the \(bx\) term with 2 terms using the numbers found in step \(1\) as coefficients of \(x\).
 Gene the starting time two terms and and so cistron the last two terms. The expressions in parentheses must be exactly the same to employ grouping.
 Factor out the expression in parentheses.
 Prepare the expressions equal to zero and solve for the variable.
Use group to factor and solve the quadratic equation: \(4x^ii+15x+ix=0\).
Solution
First, multiply \(airconditioning:4(nine)=36\). Then list the factors of \(36\).
\[1⋅36 \nonumber\]
\[2⋅18 \nonumber\]
\[3⋅12 \nonumber\]
\[4⋅9 \nonumber\]
\[6⋅6 \nonumber\]
The merely pair of factors that sums to \(fifteen\) is \(three+12\). Rewrite the equation replacing the b term, \(15x\), with two terms using \(iii\) and \(12\) as coefficients of \(x\). Gene the first ii terms, and so factor the concluding ii terms.
\[\brainstorm{align*} 4x^2+3x+12x+nine&= 0\\ x(4x+3)+three(4x+3)&= 0\\ (4x+3)(x+3)&= 0 \qquad \text{Solve using the aughtproduct belongings}\\ (4x+3)&= 3\\ ten&= \dfrac{3}{four}\\ (10+3)&= 0\\ x&= 3 \stop{marshal*}\]
The solutions are \(−\dfrac{iii}{4}\), and \(−3\). See Effigy \(\PageIndex{3}\).
Solve using factoring past group: \(12x^2+11x+2=0\).
 Respond

\((3x+2)(4x+one)=0\), \(x=−\dfrac{2}{3}\), \(x=−\dfrac{1}{4}\)
Solve the equation past factoring: \(−3x^3−5x^2−2x=0\).
Solution
This equation does not look like a quadratic, as the highest power is \(three\), not \(2\). Recall that the first thing we want to practice when solving whatever equation is to cistron out the GCF, if one exists. And it does here. We can cistron out \(−x\) from all of the terms and then keep with group.
\[\begin{align*}
3x^iii5x^ii2x&= 0\\
x(3x^2+5x+2)&= 0\\
10(3x^ii+3x+2x+2)&= 0 \qquad \text{Employ group on the expression in parentheses}\\
x[3x(x+i)+2(x+1)]&= 0\\
x(3x+2)(ten+one)&= 0\\
\text{Now, nosotros employ the cipherproduct property. Notice that nosotros take three factors.}\\
x&= 0\\
x&= 0\\
3x+2&= 0\\
x&= \dfrac{2}{3}\\
10+1&= 0\\
x&= 1
\end{align*}\]
The solutions are \(0\), \(−\dfrac{ii}{iii}\), and \(−one\).
Solve past factoring: \(x^3+11x^2+10x=0\).
 Respond

\(x=0, 10=−10, x=−one\)
Using the Square Root Property
When there is no linear term in the equation, some other method of solving a quadratic equation is by using the
square root property, in which we isolate the \(10^2\) term and have the foursquare root of the number on the other side of the equals sign. Keep in mind that sometimes we may have to manipulate the equation to isolate the \(x^ii\) term so that the square root property tin can exist used.
With the \(x^two\) term isolated, the square root belongings states that:
if \(10^2=k\), then \(x=±\sqrt{k}\)
where \(k\) is a nonzero existent number.
 Isolate the \(x^2\) term on one side of the equal sign.
 Accept the square root of both sides of the equation, putting a \(±\) sign before the expression on the side opposite the squared term.
 Simplify the numbers on the side with the \(±\) sign.
Solve the quadratic using the square root property: \(10^two=8\).
Solution
Have the square root of both sides, and then simplify the radical. Remember to use a \(±\) sign earlier the radical symbol.
\[\begin{marshal*} x^2&= eight\\ x&= \pm \sqrt{8}\\ &= \pm ii\sqrt{two} \stop{align*}\]
The solutions are \(ii\sqrt{2}\),\(two\sqrt{2}\)
Solve the quadratic equation: \(4x^2+i=seven\).
Solution
First, isolate the \(ten^ii\) term. So take the square root of both sides.
\[\brainstorm{align*} 4x^2+1&= 7\\ 4x^2&= 6\\ x^2&= \dfrac{half dozen}{4}\\ x&= \pm \dfrac{\sqrt{6}}{ii} \end{marshal*}\]
The solutions are \(\dfrac{\sqrt{6}}{ii}\), and \(\dfrac{\sqrt{6}}{2}\).
Solve the quadratic equation using the foursquare root holding: \(3{(x−4)}^2=xv\).
 Reply

\(10=4±\sqrt{5}\)
Completing the Foursquare
Not all quadratic equations tin can be factored or can be solved in their original form using the square root property. In these cases, we may utilize a method for solving a
quadratic equation
known as
completing the square. Using this method, nosotros add or subtract terms to both sides of the equation until we have a perfect square trinomial on 1 side of the equal sign. Nosotros so utilize the square root property. To complete the foursquare, the leading coefficient, \(a\), must equal \(one\). If it does not, and so divide the entire equation by \(a\). So, we can employ the following procedures to solve a quadratic equation by completing the square.
We will utilise the instance \(10^two+4x+1=0\) to illustrate each step.
Given a quadratic equation that cannot be factored, and with \(a=1\), first add or subtract the abiding term to the right sign of the equal sign.
\[\begin{marshal*}
x^2+4x+1&= 0\\
x^two+4x&= i \qquad \text{Multiply the b} \text{ term by } \dfrac{1}{2} \text{ and square information technology.}\\
\dfrac{1}{2}(4)&= 2 \\
ii^2&= iv \qquad \text{Add } \left ({\dfrac{1}{2}} \right )^ii \text{ to both sides of the equal sign and simplify the right side. We take}\\
x^2+4x+iv&= 1+four\\
x^2+4x+4&= 3 \qquad \text{The left side of the equation tin now be factored as a perfect square.}\\
{(10+ii)}^two&=iii\\
\sqrt{{(ten+2)}^2}&= \pm \sqrt{3} \qquad \text{Use the square root property and solve.}\\
\sqrt{{(x+2)}^ii}&= \pm \sqrt{iii}\\
10+ii&= \pm \sqrt{3}\\
x&= 2 \pm \sqrt{3}
\end{align*}\]
The solutions are \(−ii+\sqrt{three}\), and \(−2−\sqrt{3}\).
Solve the quadratic equation by completing the square: \(x^2−3x−v=0\).
Solution
Getgo, move the abiding term to the right side of the equal sign.
\[\begin{align*}
x^23x&= 5 \qquad \text{Then, take } \dfrac{ane}{2} \text{ of the b term and foursquare information technology.} \\
\dfrac{1}{2}(3)&= \dfrac{3}{2}\\
{\left (\dfrac{3}{ii} \correct )}^2=\dfrac{nine}{4}\\
ten^23x+{\left (\dfrac{three}{2} \right )}^two&= 5+{\left (\dfrac{3}{2} \right )}^2 \qquad \text{Add the result to both sides of the equal sign.}\\
x^ii3x+\dfrac{nine}{4}&= 5+\dfrac{9}{4}\\
\text{Factor the left side as a perfect foursquare and simplify the correct side.}\\
{\left (x\dfrac{3}{ii} \correct )}^ii&= \dfrac{29}{4}\\
(x\dfrac{3}{2})&= \pm \dfrac{\sqrt{29}}{2} \qquad \text{Use the foursquare root property and solve.}\\
x&= \dfrac{3}{2} \pm \dfrac{\sqrt{29}}{two}\\
\stop{align*}\]
The solutions are \(\dfrac{3}{2}+\dfrac{\sqrt{29}}{2}\), and \(\dfrac{3}{2}\dfrac{\sqrt{29}}{ii}\)
Solve by completing the foursquare: \(10^ii−6x=xiii\).
 Answer

\(ten=3±\sqrt{22}\)
Using the Quadratic Formula
The fourth method of solving a
quadratic equation
is by using the
quadratic formula, a formula that will solve all quadratic equations. Although the quadratic formula works on any quadratic equation in standard form, it is easy to make errors in substituting the values into the formula. Pay shut attention when substituting, and apply parentheses when inserting a negative number.
We can derive the quadratic formula by
completing the square. We volition assume that the leading coefficient is positive; if it is negative, we can multiply the equation by \(−1\) and obtain a positive a. Given \(ax^2+bx+c=0, a≠0\), nosotros will complete the square as follows:
Beginning, move the abiding term to the correct side of the equal sign:
\[ax^2+bx=−c \nonumber \]
Equally we want the leading coefficient to equal \(1\), divide through by \(a\):
\[x^ii+\dfrac{b}{a}ten=−\dfrac{c}{a} \nonumber \]
So, find \(\dfrac{1}{ii}\) of the middle term, and add together \({(\dfrac{1}{2}\dfrac{b}{a})}^ii=\dfrac{b^2}{4a^2}\) to both sides of the equal sign:
\[ten^2+\dfrac{b}{a}10+\dfrac{b^2}{4a^2}=\dfrac{b^2}{4a^ii}\dfrac{c}{a} \nonumber \]
Next, write the left side as a perfect foursquare. Detect the common denominator of the right side and write information technology as a unmarried fraction:
\[{(x+\dfrac{b}{2a})}^ii=\dfrac{b^24ac}{4a^ii} \nonumber \]
At present, use the foursquare root property, which gives
\[x+\dfrac{b}{2a}=±\sqrt{\dfrac{b^24ac}{4a^two}} \nonumber \]
\[x+\dfrac{b}{2a}=\dfrac{±\sqrt{b^ii4ac}}{2a} \nonumber \]
Finally, add \(\dfrac{b}{2a}\) to both sides of the equation and combine the terms on the right side. Thus,
\[x=\dfrac{b±\sqrt{b^24ac}}{2a} \nonumber \]
Written in standard form, \(ax^2+bx+c=0\), whatever quadratic equation can be solved using the
quadratic formula:
\[ten=\dfrac{b±\sqrt{b^24ac}}{2a}\]
where \(a\), \(b\), and \(c\) are existent numbers and \(a≠0\).
Given a quadratic equation, solve it using the quadratic formula
 Make sure the equation is in standard form: \(ax^2+bx+c=0\).
 Make note of the values of the coefficients and constant term, \(a\), \(b\), and \(c\).
 Carefully substitute the values noted in step 2 into the equation. To avoid needless errors, use parentheses around each number input into the formula.
 Calculate and solve.
Solve the quadratic equation: \(10^2+5x+i=0\).
Solution
Identify the coefficients: \(a=i,b=5,c=1\). And so use the quadratic formula.
\[\begin{align*} x&= \dfrac{(5) \pm \sqrt{(five)^twofour(1)(ane)}}{two(ane)}\\ &= \dfrac{five \pm \sqrt{254}}{two}\\ &= \dfrac{5 \pm \sqrt{21}}{2} \terminate{marshal*}\]
Employ the quadratic formula to solve \(ten^2+x+2=0\).
Solution
Offset, we identify the coefficients: \(a=1\),\(b=1\), and \(c=2\).
Substitute these values into the quadratic formula.
\[\begin{align*} ten&= \dfrac{b \pm \sqrt{b^24ac}}{2a}\\ &= \dfrac{(1) \pm \sqrt{(ane)^24(one)(2)}}{ii(1)}\\ &= \dfrac{1 \pm \sqrt{i8}}{ii}\\ &= \dfrac{ane \pm \sqrt{vii}}{ii}\\ &= \dfrac{1 \pm i\sqrt{7}}{2} \terminate{marshal*}\]
Solve the quadratic equation using the quadratic formula: \(9x^2+3x−2=0\).
 Answer

\(10=\dfrac{2}{three},ten=\dfrac{1}{3}\)
The Discriminant
The
quadratic formula
non only generates the solutions to a quadratic equation, it tells us almost the nature of the solutions when nosotros consider the
discriminant, or the expression under the radical, \(b^2−4ac\). The discriminant tells usa whether the solutions are real numbers or complex numbers, and how many solutions of each type to expect. Table \(\PageIndex{i}\) relates the value of the discriminant to the solutions of a quadratic equation.
Value of Discriminant  Results 

\(b^2−4ac=0\)  One rational solution (double solution) 
\(b^ii−4ac>0\), perfect square  Two rational solutions 
\(b^2−4ac>0\), not a perfect square  Two irrational solutions 
\(b^two−4ac<0\)  Two circuitous solutions 
For \(ax^2+bx+c=0\), where \(a\), \(b\), and \(c\) are real numbers, the
discriminant
is the expression under the radical in the quadratic formula: \(b^two−4ac\). Information technology tells united states of america whether the solutions are real numbers or circuitous numbers and how many solutions of each type to expect.
Use the discriminant to find the nature of the solutions to the postobit quadratic equations:
 \(x^2+4x+4=0\)
 \(8x^2+14x+3=0\)
 \(3x^2−5x−2=0\)
 \(3x^2−10x+15=0\)
Solution
Calculate the discriminant \(b^2−4ac\) for each equation and state the expected type of solutions.
a.
\(x^ii+4x+four=0\)
\(b^two4ac={(4)}^24(1)(4)=0\) There volition exist one rational double solution.
b.
\(8x^2+14x+3=0\)
\(b^24ac={(fourteen)}^ii4(8)(3)=100\) As \(100\) is a perfect square, there will be two rational solutions.
c.
\(3x^two−5x−2=0\)
\(b^ii4ac={(5)}^ii4(iii)(2)=49\) As \(49\) is a perfect foursquare, at that place will exist two rational solutions.
d.
\(3x^2−10x+15=0\)
\(b^24ac={(10)}^2iv(3)(xv)=80\) At that place will be ii complex solutions.
Using the Pythagorean Theorem
I of the most famous formulas in mathematics is the
Pythagorean Theorem. It is based on a right triangle, and states the human relationship among the lengths of the sides as \(a^two+b^2=c^ii\), where \(a\) and \(b\) refer to the legs of a right triangle adjacent to the \(90°\) angle, and \(c\) refers to the hypotenuse. It has immeasurable uses in compages, engineering, the sciences, geometry, trigonometry, and algebra, and in everyday applications.
We use the Pythagorean Theorem to solve for the length of one side of a triangle when we take the lengths of the other two. Because each of the terms is squared in the theorem, when nosotros are solving for a side of a triangle, we have a quadratic equation. We can use the methods for solving quadratic equations that nosotros learned in this department to solve for the missing side.
The Pythagorean Theorem is given as
\[a^2+b^2=c^2\]
where \(a\) and \(b\) refer to the legs of a correct triangle adjacent to the \(90°\) angle, and \(c\) refers to the hypotenuse, as shown in .
Find the length of the missing side of the right triangle in Figure \(\PageIndex{five}\).
Solution
As we have measurements for side \(b\) and the hypotenuse, the missing side is \(a\).
\[\begin{align*} a^two+b^two&= c^2\\ a^2+{(4)}^2&= {(12)}^ii\\ a^2+sixteen&= 144\\ a^2&= 128\\ a&= \sqrt{128}\\ &= 8\sqrt{2} \finish{align*}\]
Employ the Pythagorean Theorem to solve the right triangle trouble: Leg a measures 4 units, leg b measures 3 units. Find the length of the hypotenuse.
 Respond

\(v\) units
Access these online resources for additional instruction and practice with quadratic equations.
 Solving Quadratic Equations past Factoring
 The ZeroProduction Property
 Completing the Foursquare
 Quadratic Formula with Two Rational Solutions
 Length of a leg of a right triangle
Cardinal Equations
quadratic formula  \(x=\dfrac{−b±\sqrt{b^24ac}}{2a}\) 
Key Concepts
 Many quadratic equations can be solved by factoring when the equation has a leading coefficient of \(i\) or if the equation is a deviation of squares. The aughtfactor property is and so used to find solutions. See Example, Example, and Instance.
 Many quadratic equations with a leading coefficient other than \(ane\) can be solved past factoring using the group method. See Case and Example.
 Another method for solving quadratics is the square root property. The variable is squared. We isolate the squared term and take the foursquare root of both sides of the equation. The solution will yield a positive and negative solution. Encounter Example and Case.
 Completing the square is a method of solving quadratic equations when the equation cannot exist factored. See Example.
 A highly dependable method for solving quadratic equations is the quadratic formula, based on the coefficients and the constant term in the equation. Run across Instance.
 The discriminant is used to indicate the nature of the roots that the quadratic equation will yield: existent or circuitous, rational or irrational, and how many of each. See Example.
 The Pythagorean Theorem, amid the most famous theorems in history, is used to solve righttriangle problems and has applications in numerous fields. Solving for the length of one side of a right triangle requires solving a quadratic equation. Run across Instance.
Which Equation Can Be Used to Solve for B
Source: https://math.libretexts.org/Sandboxes/nabrao_at_chabotcollege.edu/Chabot_College_College_Algebra_for_BSTEM/02%3A_Equations_and_Inequalities/2.06%3A_Quadratic_Equations