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{{ printedBook.courseTrack.name }} {{ printedBook.name }} In standard form, quadratic equations take the form $y=ax_{2}+bx+c,$ and can be solved in various ways. In general, they are solved for the value(s) of $x$ that make the equation equal to $0.$ Thus, $y=ax_{2}+bx+c$ becomes $0=ax_{2}+bx+c.$ Graphically, all points with a $y$-coordinate of $0$ are the $x$-intercepts of the function — or the zeros of the parabola. That means, solving a quadratic equation leads to finding the zeros of the parabola. Since a parabola can have $0,1,$ or $2$ zeros, a quadratic equation can have $0,1,$ or $2$ solutions.

A quadratic equation can be solved graphically by expressing the equation as a function and identifying the solutions in the function's graph. Consider the following equation as an example.
$2x_{2}+3x−5=5x+7$ This and other quadratic equations can be solved using this method. ### 1

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 of the same degree, simplify the equation by combining like terms. $2x_{2}+3x−5−5x−7=0⇔2x_{2}−2x−12=0$

### 2

The side with the variables can now be seen as a function, $f(x).$ Plot that function in a coordinate plane. For the equation $2x_{2}−2x−12=0,$ the function is $f(x)=2x_{2}−2x−12.$

### 3

### 4

We find that the function value is $0.$ Therefore, $x=-2$ is an exact solution. The solution $x=3$ can now be verified the same way.
Since $f(3)=0$ then $x=3$ is also a solution. Thus, the equation $2x_{2}+3x−5=5x+7$ has two solutions and they are
$x=-2andx=3.$

If needed, rearrange the equation and simplify

Plot the function

Identify any points with $y$-coordinate $0$

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

Identify the $x$-coordinates

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

In the example, the $x$-coordinates are $-2$ and $3.$ Note that solving an equation graphically does not necessarily lead to an exact answer. To verify a solution, substitute it into $f(x)$ and evaluate the expression. Verifying the solution $x=-2$ is done by evaluating $f(-2).$

$f(x)=2x_{2}−2x−12$

Substitute$x=-2$

$f(-2)=2⋅(-2)_{2}−2⋅(-2)−12$

CalcPowCalculate power

$f(-2)=2⋅4−2⋅(-2)−12$

MultNegNegOnePar$-a(-b)=a⋅b$

$f(-2)=2⋅4+2⋅2−12$

MultiplyMultiply

$f(-2)=8+4−12$

AddSubTermsAdd and subtract terms

$f(-2)=0$

$f(x)=2x_{2}−2x−12$

Substitute$x=3$

$f(3)=2⋅3_{2}−2⋅3−12$

CalcPowCalculate power

$f(3)=2⋅9−2⋅3−12$

MultiplyMultiply

$f(3)=18−6−12$

AddSubTermsAdd and subtract terms

$f(3)=0$

Simple quadratic equations take the form $ax_{2}+c=0$ 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 $
Quadratic equations of the form
$ax_{2}+c=0$
can be solved using inverse operations. To undo the exponent of $2$ on the $x_{2}$ term, square roots must be used. Consider $5x_{2}−500=0.$
### 2

Now that $x_{2}$ has been isolated, it is necessary to undo the exponent. Exponents and radicals of the same index undo each other. Thus, square roots undo powers of $2$ and cube roots undo powers of $3,$ etc. To finish isolating $x_{2},$ square-root both sides of the equation.
Using a calculator to determine $100 $ gives one value — $10.$ This is because calculators give the principal root of a square root. It is necessary to remember that there are two values that, when raised to the power of $2,$ equal $100.$
$10⋅10=100and(-10)(-10)=100,$
Thus, $x=10$ and $x=-10.$

Square-root both sides of the equation

When factoring an expression in the form $ax_{2}+bx+c$ it can be difficult to ### 1

The first step is to find all possible pairs of integers that multiply to $a⋅c.$ In this case, $a=4$ and $c=3.$ Thus, their product is $3⋅4=12.$ To find all factor pairs, start with the pair where one factor is $1.$ The other factor must then be $12.$ Then continue with the pair where one of the factors is $2,$ and so forth. In this case, there are three pairs. $ 1and122and63and4 $

### 2

If the given expression is factorable, one of the factor pairs will add to equal $b.$ In this case, $b=13.$ $1+122+63+4 =13=8=7 $ Here, $1$ and $12$ is the only factor pair that adds to $13.$

### 3

Now, use the factor pair to rewrite the $x$-term of the original expression as a sum. Since the factor pair is $1$ and $12,$ the middle term can be written as $13x=12x+x.$ This gives the following equivalent expression. $4x_{2}4x_{2} +13x+12x+x +3+3 $ Notice that the expression hasn't been changed. Rather, it has been rewritten.

### 4

Now, the expression has four terms, which can be grouped into the first two terms and the last two terms. Then, the GCF of each group can be factored out. Here, begin with the first group of terms, $4x_{2}$ and $12x.$ Notice how each can be written as a product of its factors. $4x_{2}4⋅x⋅x +12x+4⋅3⋅x $ The GCF is $4x.$ Therefore, it's possible to factor out $4x.$ $4x_{2}+12x+x+34x(x+3)+x+3 $

### 5

Next, repeat the same process with the last two terms. In this case, $x$ and $3$ do not have any common factors, but it's always possible to write expressions as a product of $1$ and itself. $4x(x+3)4x(x+3) +x+3+1(x+3) $

### 6

If all previous steps have been performed correctly, there should now be two terms with a common factor, which can be seen as a repeated parentheses. Factoring $(x+3)$ out of both terms gives $4x(x+3)+(x+3)(4 1(x+3)x+1). $ This means that $4x_{2}+13x+3$ can be written in factored form as $(x+3)(4x+1).$

seeits factors. $4x_{2}+13x+3 $ This expression can be factored by finding a pair of integers whose product is $a⋅c,$ which here is $4⋅3,$ and whose sum is $b,$ which in the example is $13.$

List all pairs of integers whose product is equal to $a⋅c$

Find the pair whose sum is $b$

Write $bx$ as a sum

Factor out the GCF in the first two terms

Factor out the GCF in the last two terms

Factor out the common factor

Solve the following equation by factoring. $x_{2}−2x−3=0$

Show Solution

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+1=0x−3=0 ↔x=-1↔x=-3 $
Thus, the solutions to the equation $x_{2}−2x−3=0$ are $x=-1$ and $x=3.$

$x_{2}−2x−3=0$

$x_{2}+x−3x−3=0$

FactorOutFactor out $x$

$x(x+1)−3x−3=0$

FactorOutFactor out $-3$

$x(x+1)−3(x+1)=0$

FactorOutFactor out $(x+1)$

$(x+1)(x−3)=0$

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