5. Solving Quadratic Equations Using Square Roots
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Chapter 8
5.
Solving Quadratic Equations Using Square Roots
| Student Learning Objectives: |
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Why
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.
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.
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Challenge
Determining the Number of Solutions
Consider the following quadratic equations.
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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)
It has been shown that if d>0, the equation x^2=d has two real solutions which are sqrt(d) and - sqrt(d).
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.
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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.
He has been asked to determine the number of real solutions of three simple quadratic equations. Since Heichi only has a few minutes, he will determine the number of solutions without solving the equations. Help Heichi get ready for his trip!
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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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Pop Quiz
How Many Solutions?
Without solving the simple quadratic equations, determine the number of real solutions.
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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.
As an example, consider the equation 5x^2-500=0.
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.
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
Note that the negative solution is also considered along with the principal root when solving the equation.
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Example
Rational Solutions
Ali and Heichi are enjoying a ski vacation.
Heichi told Ali that he would pay for an extra hotel night if Ali could solve the following quadratic equation. 16x^2+15=40 Solve the equation and help Ali get an extra hotel night for free! 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 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.
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)
The equation has two real solutions and both of them are rational.
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Example
Irrational Solutions
Dominika and Magdalena are enjoying a vacation at a beach resort.
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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.
- 2x^2+7=4
x^2=3/2
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.
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
The equation has two real solutions,both of which are irrational.
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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.
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Example
Solving Other Type of Quadratic Equation
Jordan is representing North High School in an algebra competition.
She has been challenged with a quadratic equation that is a bit more complicated than a simple quadratic equation. - 2(x-5)^2+2=0 Jordan realizes that the equation can be solved by taking square roots. Help her solve the equation! Write the smallest solution first.
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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.
- 2(x-5)^2+2=0
(x-5)^2=1
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.
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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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Solving Quadratic Equations Using Square Roots
Exercise 2.1
Consider the following diagram.
The area of the inner square is 25 % the area of the bigger square. Find the side length l of the inner square.
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Let's start by finding the area of the bigger square A_b. To do so, we will use the formula for the area of a square.
Let A_i represent the area of the inner square. This area is 25 % the area of the bigger square A_b.
The area of the inner square is 49 square meters. Since l is the side length of this square, we can write an equation showing the relationship between l and 49. l^2=49 We will solve this equation by taking square roots of both sides of the equation.
The solutions to the equation are l=7 and l=- 7. However, a negative value does not make sense in this context, since a measurement of length cannot be negative. Therefore, the side length of the inner square is 7 meters.
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Hint & Answer
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Solution
Consider the following diagram.
Find the value of x.
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The triangle in the given diagram is a right triangle, so we can use the Pythagorean Theorem to write a relationship for its side lengths. a^2+b^2=c^2 Here, a and b are the legs of the triangle and c its hypotenuse. Let's substitute the values from the diagram into the above formula and simplify.
The resulting is a quadratic equation. Let's solve it by taking square roots on both sides.
Since length measurements must be positive, the negative solution does not make sense in this scenario. Therefore, the answer is x=8. Incidentally, this means that the leg lengths of the triangle are 4*8=32 units and 3*8=24 units.
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Solution
Consider the following diagram.
The area of the circle is 30 square inches. Find its radius r. Round the answer to two decimal places.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":""},"formTextBefore":"r=","formTextAfter":"inches","answer":{"text":["3.09"]}}
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Let's begin by recalling the formula for the area of a circle. A=π r^2 We are told that the area of this particular circle is 30 square inches. Let's substitute this into the formula to write a partial equation for the area of this circle. A=π r^2 ⇒ 30=π r^2 The result is a quadratic equation that can be solved by taking square roots of both sides. First we need to isolate the radius r on one side of the equation. Let's do it!
Since a radius cannot be negative, a negative solution does not make sense in this scenario. Therefore, the radius of the circle is about 3.09 inches.
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Solution
Solving Quadratic Equations Using Square Roots
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