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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| | 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
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 ax2+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.
ax2+c=0
Rule
Number of Solutions of a Simple Quadratic Equation
This type of equation can be solved using inverse operations. Once x2 is isolated, the equation can be written as x2=d, where d=-ac. The value of d gives the number of solutions the equation has.
d>0:d=0:d<0:2 real solutions1 real solution0 real 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 x2=d can be rewritten.x2=d⇔x2=d
In this case, because d>0, the expression d is a real number. Therefore, the resulting equation can be solved.
It has been shown that if d>0, the equation x2=d has two real solutions which are d and -d. d=0
If d=0, then the equation x2=d can be written as x2=0. This equation can be solved for x.x2=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 x2=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 x2=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 -4x2+5=5 will be written in the form x2=d.
In this case d is equal to 0. Therefore, the equation -4x2+5=5 has one real solution. By following a similar procedure, the other equations can be rewritten in the form x2=d.
| Equation | Rewrite as x2=d | Value of d | Number of Real Solutions |
|---|---|---|---|
| -4x2+5=5 | x2=0 | d=0 | One |
| 5x2−125=0
|
x2=25 | d=25 ⇒ d>0 | Two |
| -3x2−27=0 | x2=-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.
ax2+c=0
This type of equation can be solved using inverse operations, and two steps must be followed. - Isolate x2.
- Take square roots on both sides of the equation.
1
Isolate x2
expand_more 2
Take Square Roots
expand_more Now that x2 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.
x2=100
SqrtEqn
LHS=RHS
x2=100
a2=±a
x=±100
CalcRoot
Calculate root
x=±10
StateSol
State solutions
x= 10x=-10
Example
Rational Solutions
Ali and Heichi are enjoying a ski vacation.
16x2+15=40
Solve the equation and help Ali get an extra hotel night for free! Write the smallest solution first. {"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.58056em;vertical-align:-0.15em;\"><\/span><span class=\"mord\"><span class=\"mord mathdefault\">x<\/span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.30110799999999993em;\"><span style=\"top:-2.5500000000000003em;margin-left:0em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">1<\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.15em;\"><span><\/span><\/span><\/span><\/span><\/span><\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["-\\dfrac{5}{4}","-1.25","\\N \\dfrac{5}{4}","\\N 1.25"]}}
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Hint
Start by isolating x2.
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, x2 needs to be isolated.
Now that x2 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.
x2=1625
SqrtEqn
LHS=RHS
x2=1625
a2=±a
x=±1625
SqrtQuot
ba=ba
x=±1625
CalcRoot
Calculate root
x=±45
StateSol
State solutions
x=45x=-45(I)(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.
-2x2+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. {"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.58056em;vertical-align:-0.15em;\"><\/span><span class=\"mord\"><span class=\"mord mathdefault\">x<\/span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.30110799999999993em;\"><span style=\"top:-2.5500000000000003em;margin-left:0em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">1<\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.15em;\"><span><\/span><\/span><\/span><\/span><\/span><\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["-1.22","\\N 1.22"]}}
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Hint
Start by isolating x2.
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, x2 needs to be isolated.
Now that x2 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.
-2x2+7=4
x2=23
x2=23
SqrtEqn
LHS=RHS
x2=23
a2=±a
x=±23
StateSol
State solutions
x=23x=-23(I)(II)
(I), (II): Use a calculator
x=1.224744…x=-1.224744…
(I), (II): Round to 3 significant digit(s)
x≈1.22x≈-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.
-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. {"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.58056em;vertical-align:-0.15em;\"><\/span><span class=\"mord\"><span class=\"mord mathdefault\">x<\/span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.30110799999999993em;\"><span style=\"top:-2.5500000000000003em;margin-left:0em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">1<\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.15em;\"><span><\/span><\/span><\/span><\/span><\/span><\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["4"]}}
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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.
Closure
A Process Called Completing the Square
The quadratic equation given in the last example had a specific format.
For more information, read about the method of completing the square.
Equation: Format: -2(x−5)2+2=0 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 |
|---|---|
| -x2+4=0 | -1(x−0)2+4=0 |
| 3x2−6x+5=0 | 3(x−1)2+2=0 |
| 5x2=3x+1 | 5(x−103)2+(-2029)=0 |
| 4x2+4x=-2 | 4(x−(-21))2+1=0 |
| 4x2+4x=-2 | 4(x−(-21))2+1=0 |
| 10x2+160=80x | 10(x−4)2+0=0 |
Solving Quadratic Equations Using Square Roots
Exercises
Consider the following equation.
x2−(a−1)2=0
Here a is a constant. Find the real roots, or solutions, to this equation.
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We know that in the given quadratic equation, a is a constant. We can begin to solve this equation by isolating the x^2-term. Then we can take square roots on both sides to find its value. Remember that to find all potential solutions to the equation, we need to consider both the principal root and the negative root. Let's do it!
Therefore, the roots of the equation are x=a-1 and x=1-a.
We can check our solutions by substituting them back into the given equation and simplifying. If the result is a true statement, we know we have found the correct solutions. Let's start with x_1=a-1.
Since we ended with a true statement, we know that x_1=a-1 is a solution to the equation. Let's now check x_2=1-a.
Since we ended with another true statement, we know that x_2=1-a is also a solution to the equation.
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Solving Quadratic Equations Using Square Roots
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