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{{ printedBook.courseTrack.name }} {{ printedBook.name }} The general form of a quadratic equation is $ax_{2}+bx+c=0,$

where $a,$ $b,$ and $c$ are real numbers, and $a =0.$ There are many ways to solve such equations. One way is by using theThe quadratic formula is $x=2a-b±b_{2}−4ac ,$

where $a,b,$ and $c$ correspond with the values of a quadratic equation written in standard form, $ax_{2}+bx+c=0.$ It is derived by completing the square on the general standard form equation. The quadratic formula can be used to find solution(s) to quadratic equations.$ax_{2}+bx+c=0$

SubEqn$LHS−c=RHS−c$

$ax_{2}+bx=-c$

DivEqn$LHS/a=RHS/a$

$aax_{2}+bx =-ac $

WriteSumFracWrite as a sum of fractions

$aax_{2} +abx =-ac $

SimpQuotSimplify quotient

$x_{2}+abx =-ac $

MovePartNumRight$ca⋅b =ca ⋅b$

$x_{2}+ab ⋅x=-ac $

$x_{2}+ab ⋅x=-ac $

AddEqn$LHS+(ab /2)_{2}=RHS+(ab /2)_{2}$

$x_{2}+ab ⋅x+(ab /2)_{2}=(ab /2)_{2}−ac $

DivFracSL$ba /c=b⋅ca $

$x_{2}+ab ⋅x+(2ab )_{2}=(2ab )_{2}−ac $

$x_{2}+ab ⋅x+(2ab )_{2}=(2ab )_{2}−ac $

FacPosPerfectSquare$a_{2}+2ab+b_{2}=(a+b)_{2}$

$(x+2ab )_{2}=(2ab )_{2}−ac $

PowQuot$(ba )_{m}=b_{m}a_{m} $

$(x+2ab )_{2}=(2a)_{2}b_{2} −ac $

PowProdII$(ab)_{m}=a_{m}b_{m}$

$(x+2ab )_{2}=4a_{2}b_{2} −ac $

ExpandFrac$ba =b⋅4aa⋅4a $

$(x+2ab )_{2}=4a_{2}b_{2} −4a_{2}4ac $

SubFracSubtract fractions

$(x+2ab )_{2}=4a_{2}b_{2}−4ac $

$(x+2ab )_{2}=4a_{2}b_{2}−4ac $

SqrtEqn$LHS =RHS $

$x+2ab =±4a_{2}b_{2}−4ac $

SqrtQuot$ba =b a $

$x+2ab =±2ab_{2}−4ac $

SubEqn$LHS−2ab =RHS−2ab $

$x=-2ab ±2ab_{2}−4ac $

MoveNegFracToNumPut minus sign in numerator

$x=2a-b ±2ab_{2}−4ac $

AddSubFracAdd and subtract fractions

$x=2a-b±b_{2}−4ac $

Use the quadratic formula to solve the equation. $2x_{2}−4x−16=0$

Show Solution

Notice that the given equation is written in standard form. Thus, it can be solved using the quadratic formula. To begin, it's necessary to note the values of $a,b,$ and $c.$ It can be seen that $a=2,b=-4,andc=-16.$
To solve the equation, we can substitute these values into the formula and simplify. Remember, since a quadratic function can have $0,1,$ or $2$ roots, this equation can have $0,1,$ or $2$ real solutions.
The solutions to the equation are $x=-2$ and $x=4.$

$2x_{2}−4x−16=0$

UseQuadFormUse the Quadratic Formula: $a=2,b=-4,c=-16$

$x=2⋅2-(-4)±(-4)_{2}−4⋅2(-16) $

NegNeg$-(-a)=a$

$x=2⋅24±(-4)_{2}−4⋅2(-16) $

CalcPowProdCalculate power and product

$x=44±16+128 $

AddTermsAdd terms

$x=44±144 $

CalcRootCalculate root

$x=44±12 $

StateSolState solutions

$x=-8/4x=16/4 $

CalcQuotCalculate quotient

$x_{1}=-2x_{2}=4 $

The solutions of a quadratic equation in the form $ax_{2}+bx+c=0$ can be interpreted graphically as the zeros of the quadratic function
$y=ax_{2}+bx+c.$ If the function has **two** zeros, the equation $ax_{2}+bx+c=0$ has **two** solutions, and if the function has **one** zero, the equation has **one** solution. If the function doesn't have any zeros, the equation is said to have no real solutions.

two zeros

One zero

No real solutions

In the quadratic formula, the term under the radical sign is called the discriminant.

It's possible to use the discriminant to determine the number of solutions a quadratic equation has. $b_{2}−4acb_{2}−4acb_{2}−4ac >0⇔2real solutions=0⇔1real solution<0⇔0real solutions $ The solutions to a quadratic equation correspond to the zeros of the parabola.

Determine the number of real solutions to the equations without solving them. $x_{2}−2x+9=0andx_{2}−4x+4=0$

Show Solution

It's possible to use the discriminant of the quadratic formula to determine the number of solutions a quadratic equation has. We'll first focus on $x_{2}−2x+9=0.$ Since the equation is written in standard form, we can see that $a=1,b=-2,$ and $c=9.$ We can substitute these values into the discriminant and simplify.

$b_{2}−4ac$

SubstituteValuesSubstitute values

$(-2)_{2}−4⋅1⋅9$

CalcPowProdCalculate power and product

$4−36$

SubTermSubtract term

$-32$

$b_{2}−4ac$

SubstituteValuesSubstitute values

$(-4)_{2}−4⋅1⋅4$

CalcPowProdCalculate power and product

$16−16$

SubTermSubtract term

$0$

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