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This lesson will discuss how to solve a particular type of quadratic equation by using square roots.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
Challenge
Determining the Number of Solutions
Consider the following quadratic equations. 
It is known that a quadratic equation can have none, one, or two solutions. Can the number of solutions of each of the above equations be determined without actually solving the equations?
Discussion
Simple Quadratic Equations
A quadratic equation is a polynomial equation of degree 2. There is a special name for quadratic equations whose linear coefficient b is 0. These equations can be written in the form ax^2+c=0 and have their own characteristics.
If the linear coefficient b of a quadratic equation is 0, the equation is called a simple quadratic equation and can be written in the following form.
ax^2+c=0
Rule
Number of Solutions of a Simple Quadratic Equation
This type of equation can be solved using inverse operations. Once x^2 is isolated, the equation can be written as x^2=d, where d=- ca. The value of d gives the number of solutions the equation has.
d>0:& 2real solutions d=0:& 1real solution d<0:& 0real solutions
Proof
The cases d>0, d=0, and d<0 will be discussed one at a time.
d>0
By taking square roots, the equation x^2=d can be rewritten. x^2=d ⇔ sqrt(x^2)=sqrt(d) In this case, because d>0, the expression sqrt(d) is a real number. Therefore, the resulting equation can be solved.sqrt(x^2)=sqrt(d)
sqrt(a^2)=± a
x=± sqrt(d)
StateSol
State solutions
lcx=sqrt(d) & (I) x=- sqrt(d) & (II)
d=0
If d=0, then the equation x^2=d can be written as x^2=0. This equation can be solved for x. x^2=0 ⇔ x* x=0 By using the Zero Product Property, it can be concluded that x=0. This is the only solution for the equation.
d<0
Because the square of any real number is always greater than or equal to 0, if d<0 the equation x^2=d has no real solutions.
Example
Number of Solutions
Heichi is going on a trip with a friend. He wants to finish up his homework first, so he does not have to worry about it when he gets home.
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Hint
Write the equations in the form x^2=d. If d>0, the equation has two real solutions. If d=0, then the equation has one real solution. Finally, if d<0, the equation has no real solutions.
Solution
To start, the equation -4x^2+5=5 will be written in the form x^2=d.
In this case d is equal to 0. Therefore, the equation -4x^2+5=5 has one real solution. By following a similar procedure, the other equations can be rewritten in the form x^2=d.
| Equation | Rewrite as x^2=d | Value of d | Number of Real Solutions |
|---|---|---|---|
| -4x^2+5=5 | x^2= 0 | d= 0 | One |
| 5x^2-125=0 | x^2= 25 | d= 25 ⇒ d>0 | Two |
| - 3x^2-27=0 | x^2= - 9 | d= - 9 ⇒ d<0 | Zero |
Pop Quiz
How Many Solutions?
Without solving the simple quadratic equations, determine the number of real solutions.
Discussion
Solving Quadratic Equations with Square Roots
Apart from determining the number of real solutions of a simple quadratic equation, most of the times it is important to calculate those solutions.
Simple quadratic equations are quadratic equations whose linear coefficient b is equal to 0. 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
expand_more 2
Take Square Roots
expand_more 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
sqrt(LHS)=sqrt(RHS)
sqrt(x^2)=sqrt(100)
sqrt(a^2)=± a
x=± sqrt(100)
CalcRoot
Calculate root
x=± 10
StateSol
State solutions
lx= 10 x=- 10
Example
Rational Solutions
Ali and Heichi are enjoying a ski vacation.
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{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":" "},"formTextBefore":"x_2=","formTextAfter":null,"answer":{"text":["\\dfrac{5}{4}","1.25","\\dfrac{5}{4}","1.25"]}}
Hint
Start by isolating x^2.
Solution
In the quadratic equation, the linear coefficient is 0. Therefore, the given is a simple quadratic equation that can be solved by taking square roots. First, x^2 needs to be isolated.
Now that x^2 has been isolated, square roots can be taken on the left- and the right-hand sides. Both the principal root and the negative solution will be considered.
The equation has two real solutions and both of them are rational.
x^2=25/16
SqrtEqn
sqrt(LHS)=sqrt(RHS)
sqrt(x^2)=sqrt(25/16)
sqrt(a^2)=± a
x=± sqrt(25/16)
SqrtQuot
sqrt(a/b)=sqrt(a)/sqrt(b)
x=± sqrt(25)/sqrt(16)
CalcRoot
Calculate root
x=± 5/4
StateSol
State solutions
lcx= 54 & (I) x=- 54 & (II)
Example
Irrational Solutions
Dominika and Magdalena are enjoying a vacation at a beach resort.
Dominika told Magdalena that she would pay for an extra hotel night if Magdalena could solve the following quadratic equation.
- 2x^2+7=4
Solve the equation and help Magdalena get an extra hotel night for free! Round the solutions to three significant figures and write the smallest solution first.
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Hint
Start by isolating x^2.
Solution
In the quadratic equation, the linear coefficient is 0. Therefore, the given is a simple quadratic equation and therefore can be solved by taking square roots. First, x^2 needs to be isolated.
Now that x^2 has been isolated, square roots can be taken on the left- and the right-hand sides. Both the principal root and the negative solution will be considered.
The equation has two real solutions,both of which are irrational.
- 2x^2+7=4
x^2=3/2
x^2=3/2
SqrtEqn
sqrt(LHS)=sqrt(RHS)
sqrt(x^2)=sqrt(3/2)
sqrt(a^2)=± a
x=± sqrt(3/2)
StateSol
State solutions
lcx=sqrt(32) & (I) x=- sqrt(32) & (II)
(I), (II): Use a calculator
lx=1.224744... x=- 1.224744...
(I), (II): Round to 3 significant digit(s)
lx≈ 1.22 x≈ - 1.22
Pop Quiz
Solving Simple Quadratic Equations
Solve the following simple quadratic equations by taking square roots. If necessary, round the solutions to two decimal places.
Example
Solving Other Type of Quadratic Equation
Jordan is representing North High School in an algebra competition.
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Hint
Start by isolating (x-5)^2.
Solution
To solve the given equation, the expression (x-5)^2 will be isolated first. Then square roots will be taken on both sides.
Now that (x-5)^2 has been isolated, square roots can be taken on the left- and the right-hand sides. Both the principal root and the negative solution will be considered.
- 2(x-5)^2+2=0
(x-5)^2=1
Closure
A Process Called Completing the Square
The quadratic equation given in the last example had a specific format. Equation:& - 2(x-5)^2+2=0 Format:& a(x-h)^2+k=0 It is worth noting that all quadratic equations can be written in this format by a process called completing the square.
| Equation | Rewrite |
|---|---|
| - x^2+4=0 | - 1(x-0)^2+4=0 |
| 3x^2-6x+5=0 | 3(x-1)^2+2=0 |
| 5x^2=3x+1 | 5(x-3/10)^2+(- 29/20)=0 |
| 4x^2+4x=- 2 | 4(x-(- 1/2))^2+1=0 |
| 4x^2+4x=- 2 | 4(x-(- 1/2))^2+1=0 |
| 10x^2+160=80x | 10(x-4)^2+0=0 |
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