2. Solving Quadratics With Complex Roots
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Chapter 1
2.
Solving Quadratics With Complex Roots
| Student Learning Objectives: |
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Why
Solving quadratic equations can sometimes lead to complex and imaginary solutions. This lesson explores the intricacies of handling such equations, offering insights into the techniques used. Through various examples and scenarios, learners are introduced to the concept of complex coefficients and the challenges they present. The lesson also touches upon the use of the Quadratic Formula in the context of these equations. By understanding the underlying principles and methods, one can confidently approach and solve quadratic equations, even when they present complex and imaginary roots.
Lesson Settings & Tools
| | 11 Theory slides |
| | 10 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Solving Quadratics With Complex Roots
Slide of 11
The Quadratic Formula is a powerful tool that helps to find the real solutions to a quadratic equation, but that is not all. It is also useful for finding any complex solutions of quadratic equations. Other factoring methods can also be used to factor quadratic equations with complex roots.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
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Challenge
The Height of a Ball
One afternoon, Jordan was playing soccer in front of her building with some friends. She wanted to invite her neighbor Mark over to play. To do so, she kicked the ball upward as hard as she could, trying to get Mark to see it outside his window. That kick followed the path of the quadratic equation h=6t-t^2.
Here, h is the height in meters of the ball above the ground and t is the time in seconds after the kick. If Mark's bedroom window is 10 meters above the ground, could Mark see the ball? If so, find the value of t for the amount of time it takes for the ball to reach Mark's bedroom window. If not, select No real roots.
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Discussion
The Discriminant and the Complex Solutions
After playing with her friends, Jordan went home to do her homework. She had to find the solutions to the quadratic equation x^2-2x+2=0. Jordan knows that the Quadratic Formula is useful for finding the rational and irrational solutions to a quadratic equation. Therefore, she made a plan to use it. Quadratic Formula [0.25em] x = - b±sqrt(b^2-4 a c)/2 a She started by identifying the values of a, b, and c for the given equation. After comparing her equation with the standard form of a quadratic equation ax^2+bx+c=0, she determined that a= 1, b= -2, and c= 2. Next, she substituted these values into the Quadratic Formula.
x=- b±sqrt(b^2-4ac)/2a
▼
Substitute values and evaluate
SubstituteValues
Substitute values
x=-( -2)±sqrt(( -2)^2-4( 1)( 2))/2( 1)
CalcPowProd
Calculate power and product
x=-(-2)±sqrt(4-8)/2
NegNeg
- (- a)=a
x=2±sqrt(4-8)/2
SubTerm
Subtract term
x=2±sqrt(-4)/2
Later that evening, she was still in awe at the Quadratic Formula and wrote it on a chalkboard in her room. She then began to think about the cases in which the discriminant is negative. After analyzing the fact that there are two signs in front of the radical term, she came to another brilliant conclusion.
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Example
Pendulum Jumping
Jordan's teacher loves doing pendulum jumps off bridges and canyons. The teacher showed her students a video clip of how the jumper's trajectory repeats a path similar to a parabola after the rope is stretched. According to the teacher, her last jump can be modeled by the quadratic equation h = s^2 - 8s + 29, with 0≤ s ≤ 8.
External credits: Pseudopanax
a After the rope becomes taut, how many seconds did it take for the teacher to be 10 meters above the lake?
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c Jordan's teacher showed a second video clip that shows a jump described by the equation x^2−10x+K=0, where K is a real number. The teacher informs the students that x_1 = 5-2i is one solution to the equation. What is the other solution?
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Hint
a Substitute 10 for h in the equation modeling the jump. Before solving the equation, study the discriminant.
b Use the Quadratic Formula to solve the equation. If necessary, use complex numbers.
c If a number a+bi is a solution to a quadratic equation with real coefficients, its conjugate is also a solution. To find the conjugate, change the sign of the imaginary part.
Solution
a The distance between the jumper and the lake is given by the quadratic equation h = s^2 - 8s + 29. Since Jordan needs to determine when the jumper was 10 meters above the lake, substitute 10 for h.
10 = s^2 - 8s + 29 Group all the terms on the same side of the equation by subtracting 10 from both sides. s^2 - 8s + 19 = 0 Note that the discriminant can be evaluated before solving the equation. To do so, start by identifying the values of a, b, and c. s^2 - 8s + 19 = 0 ⇓ a= 1, b= -8, c = 19 Next, substitute these values into the expression representing the discriminant — namely, b^2-4ac.
b^2-4ac
SubstituteValues
Substitute values
( -8)^2 - 4( 1)( 19)
CalcPowProd
Calculate power and product
64 - 76
SubTerm
Subtract term
-12
Since the discriminant is negative, the equation has no real solutions. In this context, it means that the jumper never reached 10 meters above the lake after the rope became taut. In other words, the jumper was always more than 10 meters above the lake.
b Although the equation set in the previous part has no real solutions, it actually does have solutions. However, they are complex numbers, so they do not make sense in a real-life context. To find them, substitute the coefficients and the discriminant found in Part A into the Quadratic Formula.
s=- b±sqrt(b^2-4ac)/2a
▼
Substitute values and simplify
s=8±sqrt(-12)/2
SqrtNegToSqrtI
sqrt(- a)= sqrt(a)* i
s = 8± sqrt(12)i/2
▼
Simplify right-hand side
SplitIntoFactors
Split into factors
s = 8± sqrt(4* 3)i/2
SqrtProd
sqrt(a* b)=sqrt(a)*sqrt(b)
s = 8± sqrt(4)*sqrt(3)i/2
CalcRoot
Calculate root
s = 8± 2sqrt(3)i/2
FactorOut
Factor out 2
s = 2(4± sqrt(3)i)/2
SimpQuot
Simplify quotient
s = 4 ± sqrt(3)i
The solutions to the equation are s_1 = 4-sqrt(3)i and s_2 = 4+sqrt(3)i.
c Jordan's teacher gave the class an incomplete equation and one of its solutions.
Equation: & x^2-10x+K=0 Solution: & x_1 = 5-2i Jordan's friends wondered how they could find the other solution without knowing the value of K. Suddenly, Jordan remembered and shared with her friends the fact that she discovered last weekend. If a numberz=a+bi is a solution to a quadratic equation with real coefficients, then its conjugatez=a-biis also a solution. According to this information, the second solution is the conjugate of x_1 = 5-2i. To find the conjugate of a complex number, change the sign of the imaginary part. x_2 &= x_1 &⇓ x_2 &= 5-2i &⇓ x_2 &= 5+2i Therefore, the second solution is x_2 = 5+2i.
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Pop Quiz
Solving Quadratic Equations
Find the solutions to the given quadratic equation. Write the answer in the form a+bi.
Extra
How to Input the Answer- If a=0, omit it in the answer. That is, write the answer in the form bi.
- If b=1 or b=-1, write the number in the form a+i or a-i, respectively.
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Discussion
Rewriting a Sum of Squares
By now, Jordan felt quite comfortable with her handling of the Quadratic Formula and wanted to know if there were more techniques for factoring equations. Then she remembered the formula for rewriting a difference of squares. a^2-b^2 = (a+b)(a-b) She also remembered that at the time she originally learned that formula, she was a bit puzzled that there was no formula for the sum of squares. a^2+ b^2 = ? She set out to explore this formula once again. Her goal was to relate the difference of squares formula to this expression. She realized that in order to do so, she needed a minus sign between the numbers, which led her to rewrite the second term.
a^2 + b^2 = ( a+ bi)( a- bi)
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Example
Factoring a Sum of Squares
Maya is reviewing the responses to the last two-question quiz on solving quadratic equations that her math students took through the school's online platform.
a Maya started by opening Jordan's and Ali's solutions to the first question, in which they had to solve the equation x^2+9=0.
Who solved the equation correctly?
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b The second equation to solve is x^2+7=0. This time, Maya opened two windows with Jordan's and Heichi's solutions.
Who solved the equation correctly?
{"type":"choice","form":{"alts":["Only Jordan","Only Heichi","Both","Neither"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":0}
Hint
a Use the formula a^2+b^2 = (a+bi)(a-bi).
b Rewrite 7 as (sqrt(7))^2 and use the formula a^2+b^2 = (a+bi)(a-bi).
Solution
a To determine who solved the equation correctly, factor the given equation. To start, note that 9 can be rewritten as 3^2.
x^2 + 9 = 0 ⇓ x^2 + 3^2 = 0 The left-hand side expression is a sum of two squares and can be factored using complex numbers by using the following formula. a^2 + b^2 = (a+bi)(a-bi) In this case, a=x and b=3. Therefore, the given equation can be factored as follows. x^2 + 3^2 = 0 ⇓ (x+3i)(x-3i) = 0 From here, the solutions to the given equation are x_1=3i and x_2 = -3i. Consequently, Ali solved the equation correctly. In contrast, Jordan made a mistake when applying the formula. She did not rewrite 9 as 3^2 and used 9 instead of 3 for b.
b As before, factor the given equation. Note that 7 is not a perfect square and therefore cannot be rewritten as an integer squared. However, it can be written as (sqrt(7))^2.
x^2 + 7 = 0 ⇓ x^2 + (sqrt(7))^2 = 0 Once again, the left-hand side expression is the sum of two squares and can be expressed as a product of two complex numbers. a^2 + b^2 = (a+bi)(a-bi) In this case, a=x and b=sqrt(7). Consequently, the given equation can be factored as follows. x^2 + (sqrt(7))^2 = 0 ⇓ (x+sqrt(7)i)(x-sqrt(7)i)=0 The solutions to the second equation are x_1 = sqrt(7)i and x_2 = -sqrt(7)i. Comparing the computations and solutions to the responses from the students, it can be seen that Jordan answered correctly this time. In contrast, Heichi made a mistake. Because he forgot to rewrite 7 as (sqrt(7))^2, he took 7 instead of sqrt(7) for b.
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Discussion
Quadratic Equations with Complex Coefficients
After hours of homework, Jordan concluded that for a quadratic equation ax^2+bx+c=0, if b=0, then the equation can be factored using either the difference of squares formula or the sum of squares formula. x^2-y^2 &= (x+y)(x-y) [0.25em] x^2+y^2 &= (x+yi)(x-yi) Jordan noted that in both cases, the equation has two different solutions. She started wondering what would happen if the equation had only one solution. She remembered that in those cases, the quadratic equation has the form of a perfect square trinomial.
| Perfect Square Trinomial a^2 ± 2ab + b^2 = (a± b)^2 | ||
|---|---|---|
| Equation | Factorization | Solutions |
| x^2 + 6x + 9=0 | (x+3)^2=0 | x_1=x_2 = -3 |
| x^2 - 8x + 16=0 | (x-4)^2=0 | x_1=x_2 = 4 |
However, Jordan just learned that if a quadratic equation with real coefficients has one complex solution, the conjugate of that solution is also a solution to the equation. This made her come up with a good question. Can a quadratic equation have only one complex solution? For simplicity, she considered a quadratic equation that has only one pure imaginary solution. The factorization of such equation should have the form (x-bi)^2=0. Jordan realized that expanding the square would give her a clue about how the quadratic equation looks.
x^2 ± 2bix - b^2 = (x ± bi)^2
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Example
Solving Quadratic Equations with Complex Coefficients
Seeing Jordan's enthusiasm for solving quadratic equations, her older sister Dominika wanted to set her an interesting challenge.
Feeling confident, Jordan accepted the challenge.
a Dominika thought that a quadratic equation with complex coefficients would scare Jordan. For her first question, she asked Jordan to solve x^2-12ix-32 = 4.
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b Dominika wanted to make things even more complicated, so she decided to use complex coefficients that also involve radical numbers. The second equation set by Dominika is 3x^2+6sqrt(10)ix - 30 = 0.
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c As the last effort, Dominika asked for the solutions to x^2 + 4ix - 29=0.
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Hint
a Group all the constants on the left-hand side. Rewrite the quadratic expression using the fact that (x-bi)^2 = x^2-2bix - b^2.
b Divide both sides of the equation by 3. Note that 10 can be rewritten as (sqrt(10))^2. As in Part A, rewrite the quadratic expression as a perfect square trinomial with an imaginary term.
c Rewrite -29 as -4-25. Move -25 to the right-hand side and factor the left-hand side expression as a perfect square trinomial with an imaginary term. Then, solve the equation for x.
Solution
a Since Jordan cannot use the Quadratic Formula, she began by writing down the formulas to factor a quadratic equation as a perfect square trinomial with an imaginary term.
(x + bi)^2 &= x^2 + 2bix - b^2 [0.25em] (x - bi)^2 &= x^2 - 2bix - b^2 The first equation that Jordan has to solve is x^2-12ix-32=4. Since the coefficient of the x-term is negative, Jordan will use the second formula. To do so, she plans to group all the terms on the same side and rewrite each of them so that they resemble the formula.
x^2-12ix-32 = 4
SubEqn
LHS-4=RHS-4
x^2-12ix-36 = 0
SplitIntoFactors
Split into factors
x^2-2(6i)x-36 = 0
Rewrite
Rewrite 36 as 6^2
x^2-2(6i)x-6^2 = 0
Comparing the left-hand side expression to the second formula, Jordan concluded that b=6. Finally, she is ready to solve the first equation. x^2-2( 6i)x- 6^2 = 0 ⇓ (x- 6i)^2 = 0 From the resulting equation, Jordan concluded that the given equation has only one solution, 6i. Seeing how easily Jordan solved the first equation, Dominika began to worry.
b The first thing Jordan noticed about the second equation is that every coefficient is a multiple of 3. Therefore, to make her calculations easier on herself, she factored it out.
3x^2+6sqrt(10)ix -30 = 0 ⇓ [0.15em] 3(x^2+2sqrt(10)ix - 10) = 0 Next, Jordan noted that she needs to rewrite the constant term in terms of sqrt(10). She realized that 10 can be rewritten as (sqrt(10))^2. 3(x^2+2sqrt(10)ix - 10) = 0 ⇓ [0.15em] 3(x^2+2sqrt(10)ix - (sqrt(10))^2) = 0 Since the coefficient of the x-term is positive, Jordan will use the first formula written in the previous part.
3(x^2+2sqrt(10)ix - (sqrt(10))^2) = 0
DivEqn
.LHS /3.=.RHS /3.
x^2+2 sqrt(10)ix - ( sqrt(10))^2 = 0
x^2 + 2bix - b^2 = (x + bi)^2
(x + sqrt(10)i)^2 = 0
Jordan concluded that, as with the first equation, the second equation has only one solution, x=-sqrt(10)i. At this point, Dominika broke out in a cold sweat.
c The third equation written by Dominika does not seem like it can be factored as a perfect square trinomial. A little flustered, Jordan began by rewriting the coefficient of the x-term.
x^2 + 4ix - 29 = 0 ⇓ x^2 + 2(2i)x - 29 = 0 After this, Jordan noticed that the linear term is 2(2i)x implying that b=2i. Consequently, b^2 = -4. This led her to conclude that she needs to have -4 as a constant term somehow. Therefore, she decided to rewrite -29 as -4 -25. x^2 + 2(2i)x - 29 = 0 ⇓ x^2 + 2(2i)x - 4 - 25 = 0 The first three terms can be rewritten as a perfect square trinomial with an imaginary term. Then, the resulting equation can be solved for x.
x^2 + 2(2i)x - 4 - 25 = 0
AddEqn
LHS+25=RHS+25
x^2 + 2(2i)x - 4 = 25
Rewrite
Rewrite 4 as 2^2
x^2 + 2( 2i)x - 2^2 = 25
x^2 + 2bix - b^2 = (x + bi)^2
(x+ 2i)^2 = 25
SqrtEqn
sqrt(LHS)=sqrt(RHS)
x+2i = ± 5
SubEqn
LHS-2i=RHS-2i
x = ± 5 -2i
Finally, with a giant smile, Jordan concluded that the solutions to the third equation are x_1 = -5-2i and x_2 = 5-2i. Therefore, Jordan won the challenge and Dominika is not happy at all!
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Pop Quiz
Factoring Quadratic Equations with Complex Coefficients
Factor each quadratic equation.
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Closure
Finding the Height of a Ball
In the challenge presented, it was said that Jordan kicked the ball up as hard as she could, trying to make it visible from Mark's bedroom window. Mark's window is 10 meters above the ground. Additionally, it is known that the ball followed the path of the quadratic equation h=6t-t^2.
The goal is to determine whether Mark would be able to see the ball. In other words, it should be checked whether the ball can reach a height of at least 10 meters. To find this information, substitute 10 for h into the function rule that describes the path of the ball. 10 = 6t - t^2 ⇕ t^2 - 6t + 10 = 0 Next, investigate whether or not the resulting equation has real solutions by calculating the discriminant. To do so, use a= 1, b= -6, and c= 10.
b^2-4ac
▼
Substitute values and simplify
SubstituteValues
Substitute values
( -6)^2 - 4( 1)( 10)
CalcPowProd
Calculate power and product
36 - 40
SubTerm
Subtract term
-4
Extra
Solving the EquationIt can also be determined graphically if the equation 10=6t-t^2 has real solutions. To do so, graph y= 10 and y= 6t-t^2 on the same coordinate plane.
As can be seen, the graphs do not intersect, which means that the equation has no real solutions. However, the equation does have imaginary solutions, which cannot be interpreted in the given context. The Quadratic Formula can be applied to find the solutions.
10 = 6t-t^2
AddEqn
LHS+t^2-6t=RHS+t^2-6t
t^2 - 6t + 10 = 0
▼
Solve using the quadratic formula
UseQuadForm
Use the Quadratic Formula: a = 1, b= -6, c= 10
t=-( -6)±sqrt(( -6)^2-4( 1)( 10))/2( 1)
CalcPowProd
Calculate power and product
t=-(-6)±sqrt(36-40)/2
SubTerm
Subtract term
t=-(-6)±sqrt(-4)/2
NegNeg
- (- a)=a
t=6±sqrt(-4)/2
SqrtNegToSqrtI
sqrt(- a)= sqrt(a)* i
t = 6± sqrt(4)i/2
CalcRoot
Calculate root
t = 6± 2i/2
FactorOut
Factor out 2
t = 2(3± i)/2
ReduceFrac
a/b=.a /2./.b /2.
t = 3 ± i
The solutions to the quadratic equation are t_1 = 3-i and t_2 = 3+i.
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Solving Quadratics With Complex Roots
Exercise 2.1
Determine if the given statement is sometimes, always, or never true.
|
In a quadratic equation in standard form, if the signs of a and c are the same — both positive or both negative — then the solutions will be imaginary. |
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Recall that the discriminant of a quadratic equation can be used to determine the number and type of solutions to the equation. lc Quadratic equation:& ax^2+ bx+ c=0 [0.3em] Solutions:&x=- b±sqrt(b^2-4 a c)/2 a [0.8em] Discriminant:& b^2-4 a c The table below summarizes the relationship between the discriminant and the number and type of solutions.
| Discriminant (b^2-4ac) | Type and Number of Solutions |
|---|---|
| Positive | 2 real solutions |
| Zero | 1 real solution |
| Negative | 2 imaginary solutions |
Let's make some observations about the terms in the discriminant.
- The coefficient b is real, so b^2 is positive or zero.
- If a and c have the same sign, their product a c is always positive — ac>0.
- Subtracting a positive number 4ac from a non-negative number b^2 results in three scenarios — a positive number, a negative number, or zero, depending on the values of the terms.
b^2-4 a c>0 whenb^2 > 4ac [0.5em] b^2-4 a c<0 whenb^2 < 4ac [0.5em] b^2-4 a c=0 whenb^2 = 4ac If a and c have the same sign, then the discriminant can be positive, negative, or zero. Therefore, it is sometimes true that if the signs of a and c are the same — both positive or both negative — then the solutions will be imaginary.
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E
Hint & Answer
S
Solution
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Let's examine the steps that the student took to solve the quadratic equation.
To solve a quadratic equation, we first need to write all terms on one side of the equation, as the student did in Step I. After this, he had the sum of two terms 4x^2 and 5. The student then likely recalled that the sum of two perfect squares can be written as the product of two complex conjugates. a^2+b^2 ⇔ (a+bi)(a-bi) We should consider how to write each of the terms a^2=4x^2 and b^2=5 as a perfect square in order to apply this formula. 4x^2&=(2x)^2 [0.5em] 5&=(sqrt(5))^2 Note that the student did not express the number 5 as the square of a number while applying the formula Therefore, we can conclude that the expression was factored incorrectly in Step II.
In order to solve the quadratic equation, we will first write all the terms on the left-hand side and then write the terms as a sum of two perfect squares.
We will apply the formula that we expressed in Part A. To do so, we will write the sum of (2x)^2 and (sqrt(5))^2 as the product of two complex conjugates. (2x)^2+(sqrt(5))^2=0 [0.5em] ⇕ [0.5em] (2x+sqrt(5)i)(2x-sqrt(5)i)=0 We can now solve this equation by using the Zero Product Property.
We found the solution to be - sqrt(5)2i and sqrt(5)2i. Let's show the solution steps once again by correcting the student's mistake.
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Hint & Answer
S
Solution
Find all possible solutions to the equation. 2x^2+21ix-1413=95-87ix
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Let's start by writing all terms on the left-hand side of the equation by using the Properties of Equality.
Note that every coefficient is a multiple of 2. Let's factor it out! 2x^2 + 108ix - 1508 = 0 [0.15em] ⇓ [0.15em] 2(x^2 + 54ix - 754) = 0 We can simplify the equation by dividing both sides by 2. 2(x^2 + 54ix - 754) = 0 [0.15em] ⇓ [0.15em] x^2 + 54ix - 754 = 0 Note that this quadratic equation has an imaginary term. Let's recall the formulas used to factor a quadratic equation that is a perfect square trinomial with an imaginary term. (x + bi)^2 &= x^2 + 2bix - b^2 [0.25em] (x - bi)^2 &= x^2 - 2bix - b^2 Our equation does not seem like it can be factored as a perfect square trinomial. Let's rewrite the coefficient of the x-term. x^2 + 54ix - 754 = 0 ⇓ x^2 + 2(27i)x - 754 = 0 Since the x-term is 2(27i)x, b is equal to 27i. Consequently, b^2 = -729. Therefore, we need to have -729 as a constant term somehow. Let's rewrite -754 as -729 -25. x^2 + 2(27i)x - 754 = 0 ⇓ x^2 + 2(27i)x - 729 - 25 = 0 If we move the 25 to the other side of the equation, the first three terms can be rewritten as a perfect square trinomial with an imaginary term by using the first formula because the coefficient of the x-term is positive. The resulting equation can then be solved for x.
We found that the solutions to the equation are x_1 = -5-27i and x_2 = 5-27i.
We can solve quadratic equations with complex coefficients by using the Quadratic Formula. Consider the given quadratic equation. 2x^2+21ix-1413=95-87ix In the solution, we wrote all terms on the left-hand side of the equation. 2x^2 + 108ix - 1508 = 0 Then we simplified it by dividing both sides by 2, which resulted in the following equation. x^2 + 54ix - 754 = 0 We can now apply the Quadratic Formula to solve this equation. In the Quadratic Formula, b^2-4ac is the discriminant. ax^2+bx+c=0 ⇕ x=- b±sqrt(b^2-4ac)/2a Let's identify the values of a, b, and c. x^2 + 54ix - 754 = 0 ⇕ 1x^2+( 54i)x+( - 754)=0 As we can see, a= 1, b= 54i, and c= - 754. Now let's calculate the discriminant.
The discriminant is 100. We will now substitute the values of a, b, c, and the discriminant into the Quadratic Formula.
Let's write the solutions in standard form. &x_1=- 27i + 5 ⇔ x_1=5-27i &x_2=- 27i - 5 ⇔ x_2=- 5-27i The solutions of the given equation are x_1 = -5-27i and x_2 = 5-27i.
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Solving Quadratics With Complex Roots
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