5. Solving Quadratic Equations Using Square Roots
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Chapter 8
5.
Solving Quadratic Equations Using Square Roots
Understanding quadratic equations is a foundational skill in algebra. This lesson introduces learners to a specific type of quadratic equation where the linear coefficient is zero, often referred to as simple quadratic equations. By isolating the x^2 term and applying square roots to both sides, one can find the solutions or roots of the equation. The lesson emphasizes the importance of considering both the principal and negative roots to ensure all potential solutions are identified. Real-world scenarios, such as calculating the area of a rectangle, are used to demonstrate the application of these equations, making the learning process engaging and relatable.
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Lesson Settings & Tools
| | 11 Theory slides |
| | 8 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Solving Quadratic Equations Using Square Roots
Slide of 11
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
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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
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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
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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
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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 |
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How Many Solutions?
Without solving the simple quadratic equations, determine the number of real solutions.
Discussion
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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
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Rational Solutions
Ali and Heichi are enjoying a ski vacation.
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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 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
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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
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Solving Simple Quadratic Equations
Solve the following simple quadratic equations by taking square roots. If necessary, round the solutions to two decimal places.
Example
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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
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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 |
Solving Quadratic Equations Using Square Roots
Exercises
Find the real roots, or solutions, of the following quadratic equations.
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The equation can be solved by taking square roots on both sides. Remember that to ensure we find all potential solutions, we need to consider both the principal root and the negative root!
Therefore, the equation has two real solutions. These solutions are x=8 and x=-8.
This time we need to isolate the x^2-term before taking square roots on both sides. Once again, if we do not want to miss any solution, we need to consider both the principal root and the negative root.
Therefore, the equation has two real solutions. These solutions are x=3 and x=-3.
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Consider the following rectangle.
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We begin by recalling that the area of a rectangle can be found by multiplying its length l by its width w. A=l w In the diagram we can see the dimensions of the rectangle. Let's substitute them into the formula!
The area of the rectangle is 6x^2 square units.
We are told that the area of the rectangle is 54 square units. Let's substitute this value into the expression we wrote in Part A and solve for x.
Since a negative length does not make sense, we can discard the negative solution and only consider the principal root. Therefore, the answer is x=3.
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Find the real roots, or solutions, of the following quadratic equations.
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{"type":"text","form":{"type":"roots","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false,"hideNoSolution":false,"variable":"x"},"text":" "},"formTextBefore":null,"formTextAfter":null,"answer":{"texts":[]}}
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We will begin by isolating the x^2-term of the equation. Then we will take square roots on both sides. Remember that to find all of the potential solutions to the equation, we will consider both the principal root and the negative root.
Therefore, the equation has two real roots. These roots are x=2 and x=-2.
Let's see what happens if we isolate the x^2-term of the equation and take square roots on both sides.
The square root of a negative number has no real solution. This means that the equation has no real roots.
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Find the real roots, or solutions, of the following quadratic equations.
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{"type":"text","form":{"type":"roots","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false,"hideNoSolution":false,"variable":"x"},"text":" "},"formTextBefore":null,"formTextAfter":null,"answer":{"texts":["2","-6"]}}
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The given equation involves the square of a binomial. However, we do not need to expand it. Instead, we will take the square root on both sides of the equation and continue to simplify from there.
Therefore, the roots, or solutions, of the equation are x=4 and x=-2.
Let's solve this equation the same way we did in Part A. We will start by taking the square roots of both sides of the equation, then simplify as much as possible.
Therefore, the roots of the equation are x=2 and x=-6.
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Solving Quadratic Equations Using Square Roots
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